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## Vector Algebra Class 12 Important Questions with Solutions Previous Year Questions

**Algebra of Vectors**

Question 1.

Find the position vector of a point which divides the join of points with position vectors \(\vec{a}-2 \vec{b}\) and 2\(2 \vec{a}+\vec{b}\) externally in the ratio 2:1. (Delhi 2016)

Answer:

Let given position vectors are \(\overrightarrow{O A}=\vec{a}-2 \vec{b}\) and \(\overrightarrow{O B}=2 \vec{a}+\vec{b}\).

Let \(\overrightarrow{O A}\) be the position vector of a point C which divides the join of points, with position vectors

\(\overrightarrow{O A}\) and \(\overrightarrow{O B}\), externally in the ratio 2:1.

∴ \(\overrightarrow{O A}\) = \(\frac{2 \overrightarrow{O B}-1 \overrightarrow{O A}}{2-1}=\frac{2(2 \vec{a}+\vec{b})-1(\vec{a}-2 \vec{b})}{1}\) [by external section formula]

= 4\(\vec{a}\) + 2\(\vec{b}\) – \(\vec{a}\) + 2\(\vec{b}\) = 3\(\vec{a}\) + 4\(\vec{b}\)

Question 2.

If \(\vec{a}\) = 4î – ĵ + k̂ and \(\vec{b}\) = 2î – 2ĵ + k̂, then find a unit vector parallel to the vector \(\vec{a}+\vec{b}\). (All India 2016)

Answer:

Given vectors are

\(\vec{a}\) = 4î – ĵ + k̂, \(\vec{b}\) = 2î – 2ĵ + k̂.

Now, \(\vec{a}+\vec{b}\) == (4î – ĵ + k̂) + (2î – 2ĵ + k̂)

= 6î – 3ĵ + 2k̂

and \(|\vec{a}+\vec{b}|=\sqrt{(6)^{2}+(-3)^{2}+(2)^{2}}\)

= \(\sqrt{36+9+4}=\sqrt{49}\) = 7units

∴ The unit vector parallelto the vector \(\vec{a}+\vec{b}\) is

\(\frac{\vec{a}+\vec{b}}{|\vec{a}+\vec{b}|}=\frac{6 \hat{i}-3 \hat{j}+2 \hat{k}}{7}\)

Question 3.

The two vectors ĵ + k̂ and 3î – ĵ + 4k̂ represent the two sides \(\overrightarrow{A B}\) and \(\overrightarrow{A C}\) respectively of triangle ABC. Find the length of the median through A. (Delhi 2016; Foreign 2015)

Answer:

Given, \(\overrightarrow{A B}\) = ĵ + k̂ and \(\overrightarrow{A C}\) = 3î – ĵ + 4k̂

Alternate Method:

Given \(\overrightarrow{A B}\) = ĵ + k̂ and \(\overrightarrow{A C}\) = 3î – ĵ + 4k̂

Question 4.

Write the direction ratios of the vector 3\(\vec{a}\) + 2\(\vec{b}\), where \(\vec{a}\) = î + ĵ – 2k̂ and \(\vec{b}\) = 2î – 4ĵ + 5k̂ (All India 2015C)

Answer:

Clearly, 3\(\vec{a}\) + 2\(\vec{b}\) = 3 (î + ĵ – 2k̂) + 2 (2î – 4ĵ + 5k̂)

= (3î + 3ĵ – 6k̂) + (4î – 8ĵ + 10k̂)

= 7î – 5ĵ + 4k̂

Hence, direction ratios of vectors 3\(\vec{a}\) + 2\(\vec{b}\) are 7, – 5 and 4.

Question 5.

Find the unit vector in the direction of the sum of the vectors 2î + 3ĵ – k̂ and 4î – 3ĵ + 2k̂. (Foreign 2015)

Answer:

Let \(\vec{a}\) = 2î + 3ĵ – k̂ and \(\vec{b}\) = 4î – 3ĵ + 2k̂

Now, sum of two vectors,

\(\vec{a}+\vec{b}\) = (2î + 3ĵ – k̂) + (4î – 3ĵ + 2k̂) = 6î + k̂

Question 6.

Find a vector in the direction of vector 2î – 3ĵ + 6k̂ which has magnitude 21 units. (Foreign 2014)

Answer:

To find a vector in the direction of given vector, first of all we find unit vector in the direction of given vector and then multiply it with given magnitude.

Let \(\vec{a}\) = 2î – 3ĵ + 6k̂

Then, |\(\vec{a}\)| = \(\sqrt{(2)^{2}+(-3)^{2}+(6)^{2}}\)

= \(\sqrt{4+9+36}=\sqrt{49}\) = 7 units

The unit vector in the direction of the given vector \(\vec{a}\) is

Now, the vector of magnitude equal to 21 units

and in the direction of a is given by

21â = 21\(\left(\frac{2}{7} \hat{i}-\frac{3}{7} \hat{j}+\frac{6}{7} \hat{k}\right)\) = 6î – 9ĵ + 18k̂

Question 7.

Find a vector a of magnitude 5√2, making an angle of \(\frac{\pi}{4}\) with X-axis, \(\frac{\pi}{2}\) with Y-axis and an acute angle 0 with Z-axis. (All India 2014)

Answer:

Here, we have l = cos \(\frac{\pi}{4}\), m = cos \(\frac{\pi}{2}\) and n = cosθ

⇒ l = \(\frac{1}{\sqrt{2}}\), m = 0 and n = cosθ

Question 8.

Write a unit vector in the direction of the sum of the vectors \(\vec{a}\) = 2î + 2ĵ – 5k̂ and \(\vec{b}\) = -2î + ĵ – 7k̂. (Delhi2014C)

Answer:

\(\frac{1}{13}\)(4î + 3ĵ – 12k̂)

Question 9.

Find the value of p for which the vectors 3î + 2ĵ + 9k̂ and î – 2pĵ + 3k̂ are parallel. (All India 2014)

Answer:

Given, 3î + 2ĵ + 9k̂ and î – 2pĵ + 3k̂ are two parallel vectors, so their direction ratios will be proportional.

Question 10.

Write the value of cosine of the angle which the vector \(\vec{a}\) = î + ĵ + k̂ makes with Y-axis. (Delhi 2014C)

Answer:

Given, \(\vec{a}\) = î + ĵ + k̂

Now, unit vector in the direction of \(\vec{a}\) is

∴ Cosine of the angle which given vector makes with Z-axis is \(\frac{1}{\sqrt{3}}\)

Question 11.

Find the angle between X-axis and the vector î + ĵ + k̂. (All India 2014C)

Answer:

Let \(\vec{a}\) = î + ĵ + k̂

Now, unit vector in the direction of \(\vec{a}\) is

So, angle between X-axis and the vector

î + ĵ + k̂ is cos α = \(\frac{1}{\sqrt{3}}\) ⇒ α = cos^{-1} (\(\frac{1}{\sqrt{3}}\))

[∵ â = lî + mĵ + nk̂ and cos α = l ⇒ α = cos^{-1}l]

Question 12.

Write a vector in the direction of the vector î – 2ĵ + 2k̂ that has magnitude 9 units. (Delhi 2014C)

Answer:

3î – 6ĵ + 6k̂

Question 13.

Write a unit vector in the direction of vector \(\vec{P Q}\), where P and Q are the points (1, 3, 0) and (4, 5, 6), respectively. (Foreign 2014)

Answer:

First, find the vector \(\overrightarrow{P Q}\) by using the formula (x_{2} – x_{1})î + (y_{2} – y_{1})ĵ + (z_{2} – z_{1})k̂, then required unit vector is given by \(\frac{\overrightarrow{P Q}}{|\overrightarrow{P Q}|}\)

Given points are P (1, 3, 0) and Q (4, 5, 6).

Here, x_{1} = 1, y_{1} = 3, z_{1} = 0

and x_{2} = 4, y_{2} = 5, z_{2} = 6

So, vector PQ = (x_{2} – x_{1})k̂ + (y_{2} – y_{1})ĵ + (z_{2} – z_{1})k̂

= (4 – 1)î + (5 – 3)ĵ + (6 – 0)k̂

= 3î + 2ĵ + 6k̂

∴ Magnitude of given vector

Hence, the unit vector in the direction of \(\vec{P Q}\) is

Question 14.

If a unit vector \(\vec{a}\) makes angle \(\frac{\pi}{3}\) with î, \(\frac{\pi}{4}\) with ĵ and an acute angle θ with k̂, then find the value of θ. (Delhi 2013)

Answer:

Here, we have

l = cos\(\frac{\pi}{3}\), m = cos \(\frac{\pi}{4}\) and n = cos θ

Question 15.

Write a unit vector in the direction of the sum of vectors \(\vec{a}\) = 2î – ĵ + 2k̂ and \(\vec{b}\) = – î + ĵ + 3k̂. (Delhi 2013)

Answer:

\(\frac{1}{\sqrt{26}} \hat{i}+\frac{5}{\sqrt{26}} \hat{k}\)

Question 16.

If \(\vec{a}\) = xî +2ĵ – zk̂ and \(\vec{b}\) = 3î – yĵ + k̂ are two equal vectors, then write the value of x + y + z. (Delhi 2013)

Answer:

Two vectors are equal, if coefficients of their components are equal.

Given, \(\vec{a}=\vec{b}\) ⇒ xî + 2ĵ – zk̂ = î – yĵ + k̂

On comparing the coefficient of components, we get

x = 3, y = -2, z = -1

Now, x + y + z = 3 – 2 – 1 = 0

Question 17.

P and Q are two points with position vectors 3\(\vec{a}\) – 2\(\vec{b}\) and \(\vec{a}\) + \(\vec{b}\), respectively. Write the position vector of a point R which divides the line segment PQ in the ratio 2 : 1 externally. (All India 2013)

Answer:

\(-\vec{a}+4 \vec{b}\)

Question 18.

L and M are two points with position vectors 2\(\vec{a}\) – \(\vec{b}\) and \(\vec{a}\) + 2\(\vec{b}\), respectively. Write the position vector of a point N which divides the line segment LM in the ratio 2 : 1 externally. (All India 2013)

Answer:

5\(\vec{b}\)

Question 19.

A and B are two points with position vectors 2\(\vec{a}\) – 3\(\vec{b}\) and 6\(\vec{b}\) – \(\vec{a}\), respectively. Write the position vector of a point P which divides the line segment AB internally in the ratio 1:2. (All India 2013)

Answer:

Given, A and B are two points with position vectors 2\(\vec{a}\) – 3\(\vec{b}\) and 6\(\vec{b}\) – \(\vec{a}\), respectively. Also, point P divides the line segment AB in the ratio 1 : 2 internally.

Question 20.

Find the sum of the vectors \(\vec{a}\) = î – 2ĵ + k̂ \(\vec{b}\) = – 2î + 4ĵ + 5k̂ and \(\vec{c}\) = î – 6ĵ – 7k̂. (Delhi 2012)

Answer:

Given vectors are \(\vec{a}\) = î – 2ĵ + k̂ \(\vec{b}\) = – 2î + 4ĵ + 5k̂ and \(\vec{c}\) = î – 6ĵ – 7k̂.

Sum of the vectors \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) is

\(\vec{a}+\vec{b}+\vec{c}\) = (î – 2ĵ + k̂) + (- 2î + 4ĵ + 5k̂) + (î – 6ĵ – 7 k̂)

= – 4ĵ – k̂

Question 21.

Find the sum of the following vectors. \(\vec{a}\) = î – 3k̂, \(\vec{b}\) = 2ĵ – k̂, \(\vec{c}\) = 2î – 3ĵ + 2k̂. (Delhi 2012)

Answer:

3î – ĵ – 2k̂

Question 22.

Find the sum of the following vectors. \(\vec{a}\) = î – 2ĵ, \(\vec{b}\) = 2î – 3 ĵ, \(\vec{c}\) = 2î + 3k̂. (Deihi 2012)

Answer:

5î – 5ĵ + 3k̂

Question 23.

Find the scalar components of \(\vec{AB}\) with initial point A (2,1) and terminal point B(- 5, 7). (All India 2012)

Answer:

Given initial point is A (2,1) and terminal point is B (- 5, 7), then scalar component of \(\overrightarrow{A B}\) are

x_{2} – x_{1} = – 5 – 2 = – 7and y_{2} – y_{1} = 7 – 1 = 6.

Question 24.

For what values of \(\vec{a}\), the vectors 2î – 3ĵ + 4k̂ and aî + 6ĵ – 8k̂ are collinear? (Delhi 2011)

Answer:

If \(\vec{a}\) and \(\vec{b}\) are collinear, then use the condition \(\vec{a}\) = λ\(\vec{b}\), where λ is some scalar.

Let given vectors are \(\vec{a}\) = 2î – 3ĵ + 4k̂ and \(\vec{a}\) = aî + 6ĵ – 8k̂

We know that, vectors \(\vec{a}\) and \(\vec{b}\) are said to be collinear, if

\(\vec{a}\) = k. \(\vec{b}\), where k is a scalar.

∴ 2î – 3ĵ + 4k̂ = k(aî + 6ĵ – 8k̂)

On comparing the coefficients of î and ĵ, we get

2 = ka and -3 = 6k ⇒ k = –\(\frac{1}{2}\)

∴ 2 = –\(\frac{1}{2}\)a ⇒ a = -4

Question 25.

Write the direction cosines of vector -2î + ĵ – 5k̂. (Delhi 2011)

Answer:

Direction cosines of the vector aî + bĵ + ck̂ are

Question 26.

Write the position vector of mid-point of the vector joining points P(2, 3, 4) and Q (4, 1, – 2). (Foreign 2011)

Answer:

Mid-point of the position vectors

\(\vec{a}\) = a_{1}î + a_{2}ĵ + a_{3}k̂ and

\(\vec{b}\) = b_{1}î + b_{2}ĵ + b_{3}k̂ is \(\frac{\vec{a}+\vec{b}}{2}\) or \(\frac{\left(a_{1}+b_{1}\right) \hat{i}+\left(a_{2}+b_{2}\right) \hat{j}+\left(a_{3}+b_{3}\right) \hat{k}}{2}\)

Given points are P(2, 3, 4) and Q(4,1,-2) whose position vectors are \(\overrightarrow{O P}\) = 2 î + 5ĵ + 4k̂ and \(\overrightarrow{O Q}\) =4î + ĵ – 2k̂.

Now, position vector of mid-point of vector joining points P(2, 3, 4) and Q(4, 1, – 2) is

Question 27.

Write a unit vector in the direction of vector \(\vec{a}\) = 2î + ĵ + 2k̂. (All India 2011; Delhi 2009)

Answer:

We know that, unit vector in the direction of â is â = \(\frac{\vec{a}}{|\vec{a}|}\)

Required unit vector in the direction of vector

\(\vec{a}\) = 2î + ĵ + 2k̂

Question 28.

Find the magnitude of the vector \(\vec{a}\) = 3î – 2ĵ + 6k̂. (All India 2011C: Delhi 2008)

Answer:

Magnitude of a vector r = xî + yĵ + zk̂ is |\(\vec{r}\)| = \(\sqrt{x^{2}+y^{2}+z^{2}}\)

Given vector is a = 3i – 2/ + 6fc.

∴ Magnitude of \(\vec{a}\) = \(|\vec{a}|=\sqrt{(3)^{2}+(-2)^{2}+(6)^{2}}\)

= \(\sqrt{9+4+36}=\sqrt{49}\) = 7 units

Question 29.

Find a unit vector in the direction of vector \(\vec{a}\) = 2î + 3ĵ + 6k̂. (All India 2011C)

Answer:

\(\frac{2}{7} \hat{i}+\frac{3}{7} \hat{j}+\frac{6}{7} \hat{k}\)

Question 30.

If A, B and C are the vertices of a ΔABC, then what is the value of \(\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{C A}\) ? (Delhi 2011C)

Answer:

Let ΔABC be the given triangle.

Now, by triangle law of vector addition,

Question 31.

Find a unit vector in the direction of \(\vec{a}\) = 2î – 3ĵ + 6k̂. (Delhi 2011c)

Answer:

\(\frac{2}{7} \hat{i}-\frac{3}{7} \hat{j}+\frac{6}{7} \hat{k}\)

Question 32.

Find a vector in the direction of \(\vec{a}\) = 2î – ĵ + 2k̂, which has magnitude 6 units. (Delhi 2010C)

Answer:

4î – 2ĵ + 4k̂

Question 33.

Find the position vector of mid-point of the line segment AB, where A is point (3, 4, -2) and Bis point (1, 2, 4). (Delhi 2010)

Answer:

2î + 3ĵ + k̂

Question 34.

Write a vector of magnitude 9 units in the direction of vector -2î + ĵ + 2k̂. (All India 2010)

Answer:

-6î + 3ĵ + 6k̂

Question 35.

Write a vector of magnitude 15 units in the direction of vector î – 2ĵ + 2k̂. (Delhi 2010)

Answer:

5î – 10ĵ + 10k̂

Question 36.

What is the cosine of angle which the vector √2î + ĵ + k̂ makes with Y-axis? (Delhi 2010)

Answer:

\(\frac{1}{2}\)

Question 37.

Find a vector of magnitude 5 units and parallel to the resultant of \(\vec{a}\) = 2î + 3ĵ – k̂ and \(\vec{b}\) = î – 2ĵ + k̂. (Delhi 2011)

Answer:

First, find resultant of the vectors a and o, which is \(\vec{a}\) + \(\vec{b}\). Then, find a unit vector in the direction of \(\vec{a}\) + \(\vec{b}\). After this, the unit vector is multiplying by 5.

Given, \(\vec{a}\) = 2î + 3ĵ – k̂ and \(\vec{b}\) = î – 2ĵ + k̂.

Now, resultant of above vectors = \(\vec{a}\) + \(\vec{b}\)

= (2î + 3ĵ – k̂) + (î – 2ĵ + k̂) = 3î + ĵ

Question 38.

Let \(\vec{a}\) = î + ĵ + k̂, \(\vec{b}\) = 4î – 2ĵ + 8k̂ and \(\vec{c}\) = î – 2ĵ + k̂. Find a vector of magnitude 6 units, which is parallel to the vector 2\(\vec{a}\) – \(\vec{b}\) + 8 \(\vec{c}\). (All India 2010)

Answer:

First, find the vector 2 a – b + 3c, then find a unit vector in the direction of 2a-b + 3c.

After this, the unit vector is multiplying by 6.

Given, \(\vec{a}\) = î + ĵ + k̂, \(\vec{b}\) = 4î – 2ĵ + 8k̂ and \(\vec{c}\) = î – 2ĵ + k̂

∴ \(2 \vec{a}-\vec{b}+3 \vec{c}\)

= 2 (î + ĵ + k̂) – (4î – 2ĵ + 3k̂) + 3 (î – 2ĵ + k̂)

= 2î + 2ĵ + 2k̂ – 4î + 2ĵ – 3k̂ + 3î – 6ĵ + 3k̂

⇒ \(2 \vec{a}-\vec{b}+3 \vec{c}\) = î – 2ĵ + 2k̂

Now, a unit vector in the direction of vector

Hence, vector of magnitude 6 units parallel to the Vector \(2 \vec{a}-\vec{b}+3 \vec{c}\) = 6\(\left(\frac{1}{3} \hat{i}-\frac{2}{3} \hat{j}+\frac{2}{3} \hat{k}\right)\)

= 2î – 4ĵ + 4k̂

Question 39.

Find the position vector of a point R, which divides the line joining two points P and Q whose position vectors are 2\(\vec{a}\) + \(\vec{b}\) and \(\vec{a}\) – 8\(\vec{b}\) respectively, externally in the ratio 1 : 2. Also, show that P is the mid-point of line segment RO. (Delhi 2010)

Answer:

Given, \(\overrightarrow{O P}\) = Position vector of P = 2\(\vec{a}\) + \(\vec{b}\)

and \(\overrightarrow{O Q}\) = Position vector of Q = \(\vec{a}\) – 3\(\vec{b}\)

Let OR be the position vector of point R, which divides PQ in the ratio 1 : 2 externally

Now, we have to show that P is the mid-point of RQ,

Hence, P is the mid-point of line segment RQ.

**Product of Two Vectors and Scalar Triple Product**

Question 1.

Find the magnitude of each of the two vectors \(\vec{a}\) and \(\vec{b}\), having the same magnitude such that the angle between them is 60° and their scalar product is \(\frac{9}{2}\). (CBSE 2018)

Answer:

Question 2.

Find the value of [î, k̂, ĵ], (CBSE 2018C)

Answer:

[î, k̂, ĵ] = î ∙ (k̂× ĵ)

= -[[î, k̂, ĵ] = –\(\left|\begin{array}{lll}

1 & 0 & 0 \\

0 & 1 & 0 \\

0 & 0 & 1

\end{array}\right|\) = – 1

Question 3.

Find λ and μ, if (î + 3ĵ + 9k̂) × (3î – λĵ + μk̂) = 0. (All India 2016)

Answer:

Given, (î + 3ĵ + 9k̂) × (3î – λĵ + μk̂)

= î(3μ + 9λ) ĵ k̂

On comparing the coefficients of î, ĵ and k̂ , we get

3μ + 9λ = 0, – μ + 27 = 0 and – λ – 9 = 0

⇒ μ = 27 and – λ = 9

⇒ μ = 27 and λ = – 9

Also, the values of μ and λ satisfy the equation

3μ + 9λ = 0.

Hence, μ = 27 and λ = – 9.

Question 4.

Write the number of vectors of unit length perpendicular to both the vectors \(\vec{a}\) = 2î + ĵ + 2k̂ and \(\vec{b}\) = ĵ + k̂. (All India 2016)

Answer:

We know that, unit vectors perpendicular to \(\vec{a}\)

and \(\vec{b}\) are ±\(\left(\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}\right)\)

So, there arc two unit vectors perpendicular to the given vectors.

Question 5.

If \(\vec{a}, \vec{b}, \vec{c}\) are unit vectors such that \(\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}\) = 0, then write the value of \(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}\). (Foreign 2016)

Answer:

Question 6.

If \(|\vec{a} \times \vec{b}|^{2}+|\vec{a} \cdot \vec{b}|^{2}\) = 400 and \(|\vec{a}|\) = 5, then write the value of \(|\vec{b}|\). (Foreign 2016)

Answer:

Question 7.

Find λ, if the vectors \(\vec{a}\) = î + 3ĵ + k̂, \(\vec{b}\) = 2 î – ĵ – k̂ and \(\vec{c}\) = λĵ + 3 k̂ are coplanar. (Delhi 2015)

Answer:

⇒ 1(- 3 + λ) – 3(6) + 1(2λ) = 0

[expanding along R_{1}]

⇒ – 3 + λ – 18 + 2λ = 0

⇒ 3λ = 21

∴ λ = 7

Question 8.

If \(\vec{a}\) = 7î + ĵ – 4k̂ and \(\vec{b}\) = 2î + 6ĵ + 3k̂, then find the projection of \(\vec{a}\) on \(\vec{b}\). (Delhi 2015)

Answer:

Question 9.

If â, b̂ and ĉ are mutually perpendicular unit vectors, then find the value of |2â + b̂ + ĉ |. (All India 2015)

Answer:

Given â, b̂ and ĉ are mutually perpendicular unit vectors, i.e.

Question 10.

Write a unit vector perpendicular to both the vectors \(\vec{a}\) = î + ĵ + k̂ and \(\vec{b}\) = î + ĵ. (All India 2015)

Answer:

First, determine perpendicular vectors of \(\vec{a}\) and \(\vec{b}\), i.e., \(\vec{a} \times \vec{b}\). Further , determine perpendicular unit vector by using formula \(\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}\).

Given vector are \(\vec{a}\) = î + ĵ + k̂ and \(\vec{b}\) = î + ĵ

As we know the, vectors \(\vec{a} \times \vec{b}\) is perpendicular to both the vectors, so let us first evaluate \(\vec{a} \times \vec{b}\).

Then, \(\vec{a} \times \vec{b}\) = \(\left|\begin{array}{lll}

\hat{i} & \hat{j} & \hat{k} \\

1 & 1 & 1 \\

1 & 1 & 0

\end{array}\right|\)

= î(0 -1) – ĵ(0 – 1) + k̂(1 – 1)

= – î + ĵ

Then , the unit vector perpendicular to both \(\vec{a}\) and \(\vec{b}\) is given by

Question 11.

Find the area of a parallelogram whose adjacent sides are represented by the vectors 2 î – 3 k̂ and 4 ĵ + 2 k̂. (Foreign 2015)

Answer:

Let adjacent sides of a parallelogram bc

\(\vec{a}\) = 2 î – 3 k̂ and \(\vec{b}\) = 4 ĵ + 2 k̂.

Question 12.

If \(\vec{a}\) and \(\vec{b}\) are perpendicular vectors, |\(\vec{a}\) + \(\vec{b}\)| = 13 and |\(\vec{a}\)| = 5, then find the value of |\(\vec{b}\)|. (All India 2014)

Answer:

Question 13.

If \(\vec{a}\) and \(\vec{b}\) are two unit vectors such that \(\vec{a}\) + \(\vec{b}\) is also a unit vector, then find the angle between \(\vec{a}\) and \(\vec{b}\). (Delhi 2014)

Answer:

Question 14.

Find the projection of the vector î + 3ĵ + 7k̂ on the vector 2î – 3 ĵ + 6k̂. (Delhi 2014)

Answer:

let \(\vec{a}\) = î + 3ĵ + 7k̂ and \(\vec{a}\) = 2î – 3 ĵ + 6k̂

Question 15.

Write the projection of vector î + ĵ + k̂ along the vector ĵ. (Foreign 2014)

Answer:

1

Question 16.

Write the value of the following. î × (ĵ + k̂) + ĵ × (k̂ + î) + k̂ × (î + ĵ). (Foreign 2014)

Answer:

we have, î × (ĵ + k̂) + ĵ × (k̂ + î) + k̂ × (î + ĵ)

= î × ĵ + î × k̂ × ĵ × k̂ + ĵ × î + k̂ × î + k̂ × ĵ

[∵ cross product is distributive over addition]

= k̂ – ĵ + î – k̂ + ĵ – î = \(\vec{0}\)

[∵ î × ĵ = k̂, î × k̂ = – ĵ, ĵ × k̂ = î, ĵ × î = – k̂, k̂ × î = ĵ, k̂ × ĵ = – î ]

Question 17.

If vectors \(\vec{a}\) and \(\vec{b}\) are such that |\(\vec{a}\)| = 3, |\(\vec{b}\)| = 2/3 and \(\vec{a}\) × \(\vec{b}\) is a unit vector, then write the angle between \(\vec{a}\) and \(\vec{b}\). (Delhi 2014: All India 2010)

Answer:

Question 18.

Find \(\vec{a} \cdot(\vec{b} \times \vec{c})\), if \(\vec{a}\) = 2î + ĵ + 3k̂, \(\vec{b}\) = -î + 2ĵ + k̂, and \(\vec{c}\) = 3î + ĵ + 2k̂. (All India 2014)

Answer:

Given, \(\vec{a}\) = 2î + ĵ + 3k̂, \(\vec{b}\) = -î + 2ĵ + k̂, and \(\vec{c}\) = 3î + ĵ + 2k̂.

= 2(4 – 1) – 1 (- 2 – 3) + 3( – 1 – 6)

= 2 × 3 – 1 × (-5) + 3 × (- 7)

= 6 + 5 – 21 = 11 – 21 = – 10

Question 19.

If \(\vec{a}\) and \(\vec{b}\) are unit vectors, then find the angle between \(\vec{a}\) and \(\vec{b}\), given that (√3\(\vec{a}\) – \(\vec{b}\)) is a unit vector. (Delhi 2014C)

Answer:

Question 20.

If |\(\vec{a}\)| = 8, |\(\vec{b}\)| = 3 and|\(|\vec{a} \times \vec{b}|\)| = 12, find the angle between \(\vec{a}\) and \(\vec{b}\). (All India 2014C)

Answer:

let θ be the angle between latex]\vec{a}[/latex] and \(\vec{b}\).

Question 21.

Write the projection of the vector \(\vec{a}\) = 2 î – ĵ + k̂ on the vector \(\vec{b}\) = î + 2ĵ + 2k̂. (Delhi 2014 C)

Answer:

\(\frac{2}{3}\)

Question 22.

Write the value of λ, so that the vectors a = 2î + λĵ + k̂ and b = î – 2ĵ + 3k̂ are perpendicular to each other. (Delhi 2013C, 2008)

Answer:

Given vectors are \(\vec{a}\) = 2î + λĵ + k̂

and \(\vec{b}\) = î – 2ĵ + 3k̂

Since, vectors are perpendicular.

∴ \(\vec{a} \cdot \vec{b}\) = 0

⇒ (2î + λĵ + k̂) ∙ (î – 2ĵ + 3k̂)

⇒ 2 – 2λ + 3 = 0

∴ λ = 5/2

Question 23.

Write the projection of (\(\vec{b}\) + \(\vec{c}\)) on \(\vec{a}\), where \(\vec{a}\) = 2î – 2ĵ + k̂, \(\vec{b}\) = î + 2ĵ – 2k̂ and \(\vec{c}\) = 2î – ĵ + 4k̂. (All India 2013 C)

Answer:

Question 24.

Write the projection of the vector 7î + ĵ – 4k̂ on the vector 2î + 6 ĵ + 3k̂. (Delhi 2013C)

Answer:

\(\frac{8}{7}\)

Question 25.

If \(\vec{a}\) and \(\vec{b}\) are two vectors such that |\(\vec{a}\) + \(\vec{b}\)| = |\(\vec{a}\)|, then prove that vector 2\(\vec{a}\) + \(\vec{b}\) is perpendicular to vector b. (Delhi 2013)

Answer:

Question 26.

Find |\(\vec{x}\)|, if for â unit vector a, \((\vec{x}-\vec{a}) \cdot(\vec{x}+\vec{a})\) = 15. (All India 2013)

Answer:

Given, \(\vec{a}\) is a unit vector. Then, |\(\vec{a}\)| = 1.

Question 27.

Find λ, when projection of \(\vec{a}\) = λî + ĵ + 4k̂ on \(\vec{b}\) = 2 î + 6 ĵ + 3k̂ is 4 units. (Delhi 2012)

Answer:

Given, \(\vec{a}\) = λî + ĵ + 4k̂ on \(\vec{b}\) = 2 î + 6 ĵ + 3k̂ and projection of \(\vec{a}\) and \(\vec{b}\) = 4.

⇒ 2λ + 18 = 28

⇒ 2λ = 10

∴ λ = 5

Question 28.

Write the value of (k̂ × ĵ) . î + ĵ . k̂. (All India 2012)

Answer:

Use the results k̂ × ĵ = – î

ĵ ∙ k̂ and î ∙ î = 1 and simplify it.

Given, (k̂ × î) ∙ î + ĵ ∙ k̂ = (- î) ∙ î + ĵ ∙ k̂

= – (î ∙ î) + 0 = – 1 [∵ (î ∙ î) = 1]

Question 29.

If \(\vec{a} \cdot \vec{a}\) = 0 and \(\vec{a} \cdot \vec{b}\) = 0, then what can be concluded about the vector \(\vec{b}\)? (Foreign 2011)

Answer:

From Eqs. (i) and (ii). it may be concluded that \(\vec{b}\) is either zero or non-zero perpendicular vector.

Question 30.

Write the projection of vector î – ĵ on the vector î + ĵ. (All India 2011)

Answer:

0

Question 31.

Write the angle between vectors \(\vec{a}\) and \(\vec{b}\) with magnitudes √3 and 2 respectively, having \(\vec{a}\). \(\vec{b}\) = √6. (All India 2011)

Answer:

let θ be the angle between \(\vec{a}\) and \(\vec{b}\), then use the following formula

cos θ = \(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}\).

Question 32.

For what value of λ are the vectors î + 2λĵ + k̂ and 2î + ĵ – 3k̂ perpendicular? (All India 2011C)

Answer:

\(\frac{1}{2}\)

Question 33.

If |\(\vec{a}\)| = √3, |\(\vec{b}\)| = 2 and angle between \(\vec{a}\) and \(\vec{b}\) is 60°, then find \(\vec{a}\).\(\vec{b}\). (Delhi 2011C)

Answer:

Question 34.

Find the value of λ, if the vectors 2î + λĵ + 3k and 3î + 2ĵ – 4k̂ are perpendicular to each other. (All India 2010c)

Answer:

3

Question 35.

If |\(\vec{a}\)| = 2, |\(\vec{b}\)| = 3 and \(\vec{a}\).\(\vec{b}\) = 3, then find the projection of \(\vec{b}\) on \(\vec{a}\). (All India 2010C)

Answer:

Question 36.

If \(\vec{a}\) and \(\vec{b}\) are two vectors, such that \(|\vec{a} \cdot \vec{b}|=|\vec{a} \times \vec{b}|\), then find the angle between \(\vec{a}\) and \(\vec{b}\). (All India 2010)

Answer:

Use the following formulae:

\(\vec{a} \cdot \vec{b}\) = \(|\vec{a}||\vec{b}|\) cos θ

and \(|\vec{a} \times \vec{b}|\) = \(|\vec{a}||\vec{b}|\) sin θ

where, θ is the angle between \(\vec{a}\) and \(\vec{b}\).

Question 37.

Find λ, if (2î + 6ĵ + 14k̂) × (î – λĵ + Ik̂) = 0. (All India 2010)

Answer:

– 3

Question 38.

If the sum of two unit vectors a and b is a unit vector, show that the magnitude of their difference is √3. (Delhi 2019, 2012c)

Answer:

let \(\vec{c}\) = \(\vec{a}\) + \(\vec{b}\). Then, according to given condition \(\vec{c}\) is a unit vector, i.e. |\(\vec{c}\)| = 1.

[taking positive square root, as magnitude cannot be negative]

Question 39.

If \(\vec{a}\) = 2î + 5ĵ + k̂, \(\vec{b}\) = î – 2 ĵ + k̂ and \(\vec{c}\) = – 3î + ĵ + 2k̂, find \([\vec{a} \vec{b} \vec{c}]\). (Delhi 2019)

Answer:

= 2(- 4 – 1) – 3(2 + 3) + 1(1 – 6)

= – 10 – 15 – 5 = – 30

Question 40.

If |\(\vec{a}\)| = 2, |\(\vec{b}\)| = 7 and \(\vec{a} \times \vec{b}\) = 3î + 2ĵ + 6k̂, find the angle between \(\vec{a}\) and \(\vec{b}\). (All India 2019)

Answer:

let θ be the angle between \(\vec{a}\) and \(\vec{b}\).

Question 41.

Find the volume of cuboid whose edges are given by -5î + 7ĵ + 5k̂, -5î + 7ĵ – 5k̂ and 7î – 5 ĵ – 5k̂. (All India 2019)

Answer:

= |- 3 (- 21 – 15) – 7 (15 + 21) + 5(25 – 49)|

= |1108 – 252 – 120|

= 264 cubic units

Question 42.

Show that the points A(-2î + 5ĵ + 5k̂), B(î + 2 ĵ + 5k̂) and C(7î – k̂) are collinear. (All India 2019)

Answer:

Question 43.

Find \(|\vec{a} \times \vec{b}|\), if \(\vec{a}\) = 2î + ĵ + 5k̂ and \(\vec{b}\) = 3î + 5ĵ – 2k̂. (All India 2019)

Answer:

We have, \(\vec{a}\) = 2î + ĵ + 3k̂ and \(\vec{b}\) = 3î + 5ĵ – 2k̂

∴ \(\vec{a} \times \vec{b}\) = \(\left|\begin{array}{rrr}

\hat{i} & \hat{j} & \hat{k} \\

2 & 1 & 3 \\

3 & 5 & -2

\end{array}\right|\)

= î ( – 2 – 15) – ĵ (- 4 – 9) + k̂(10 – 3)

= – 17î + 13ĵ + 7k̂

Question 44.

If θ is the angle between two vectors î – 2 ĵ + 3k̂ and 3î – 2 ĵ + k̂, find sin θ. (CBSE 2018)

Answer:

let \(\vec{a}\) = î – 2 ĵ + 3k̂ and \(\vec{b}\) 3î – 2 ĵ + k̂

Question 45.

If \(\vec{a}+\vec{b}+\vec{c}\) = 0 and |\(\vec{a}\)| = 5, |\(\vec{b}\)| = 6 and |\(\vec{c}\)| = 9, then find the angle between \(\vec{a}\) and \(\vec{b}\). (CBSE 2018C)

Answer:

(5)^{2} + 2 × 5 × 6 × cos θ + (6)^{2} = (9)^{2}

⇒ 25 + 60 cos θ + 36 = 81

⇒ cos θ = \(\frac{20}{60}=\frac{1}{3}\)

⇒ θ = cos^{-1}\(\left(\frac{1}{3}\right)\)

Question 46.

If î + ĵ + k̂, 2î + 5ĵ, 5î + 2ĵ – 5k̂ and î – 6ĵ – k̂ respectively, are the position vectors of points A, B, C and D, then find the angle between the straight lines AB and CD. Find whether \(\overrightarrow{A B}\) and \(\overrightarrow{C D}\) are collinear or not. (Delhi 2019)

Answer:

Question 47.

The scalar product of the vector \(\vec{a}\) = î + ĵ + k̂ with a unit vector along the sum of the vectors \(\vec{b}\) = 2î + 4ĵ – 5k̂ and \(\vec{c}\) = λî + 2ĵ + 5k̂ is equal to 1. Find the value of λ and hence find the unit vector along \(\vec{b}\) + \(\vec{c}\). (All India 2019)

Answer:

Question 48.

Let \(\vec{a}\) = 4 î + 5ĵ – k̂, \(\vec{b}\) = î – 4ĵ + 5k̂ and \(\vec{c}\) = 3î + ĵ – k̂. Find a vector which is perpendicular to both \(\vec{c}\) and \(\vec{b}\) and \(\vec{d} \cdot \vec{a}\) = 21. (CBSE 2018)

Answer:

We have, \(\vec{a}\) = 4 î + 5ĵ – k̂, \(\vec{b}\) = î – 4ĵ + 5k̂ and \(\vec{c}\) = 3î + ĵ – k̂.

Since, \(\vec{d}\) is perpendicular to both \(\vec{c}\) and \(\vec{b}\).

= λ[î(5 – 4) – ĵ(15 + 1) + k̂(- 12 – 1)]

= λ(î – 16ĵ – 13k̂)

Also, it is given that \(\vec{a} \cdot \vec{a}\) = 21

∴ λ(î – 16ĵ – 13k̂) ∙ (4î + 5ĵ – k̂) = 21

⇒ λ(4 – 80 + 13) = 21

⇒ λ(- 63) = 21

⇒ λ = \(\frac{-1}{3}\)

Now from Eq. (j), we get

\(\vec{d}\) = –\(\frac{-1}{3}\)(î – 16ĵ – 13k̂)

Question 49.

Find x such that the four points A(4, 4, 4), B(5, x, 8), C(5, 4, 1) and D (7, 7, 2) are coplanar. (CBSE 2018C)

Answer:

Given points are A(4, 4, 4), B (5, x, 8), C(5, 4, 1) and D(7, 7, 2), then position vectors of A, B, C and D respectively, are

⇒ 1(0 + 9) – (x – 4) (- 2 + 9) + 4(3 – 0) = 0

⇒ 9 – (x – 4) (7) + 12 = 0

⇒ 9 – 7x + 28 + 12 = 0

⇒ 49 – 7x = 0

⇒ 7x = 49

⇒ x = 7

Question 50.

Find the value of x such that the points A(3, 2, 1), B(4, x, 5), C(4, 2,- 2) and D (6, 5, -1) are coplanar. (All India 2017)

Answer:

5

Question 51.

If \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are three mutually perpendicular vectors of the same magnitude, then prove that \(\vec{a}+\vec{b}+\vec{c}\) is equally inclined with the vectors \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\). (Delhi 2017, 2013C, 2011)

Answer:

If three vectors \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are mutually perpendicular to each other, then \(\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{c}\) = \(\vec{c} \cdot \vec{a}\) = 0 and if all three vectors \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are equally inclined with the vector \((\vec{a}+\vec{b}+\vec{c})\) that means each vector \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) makes equal angle with \((\vec{a}+\vec{b}+\vec{c})\) by using formula

cos θ = \(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}\).

Question 52.

Using vectors, find the area of the ΔABC, whose vertices are A(1, 2, 5), 5(2, -1, 4) and C(4, 5, -1). (Delhi 2017; All India 2013)

Answer:

Let the position vectors of the verices A, B and C of ΔABC be

Question 53.

Let \(\vec{a}\) = î + ĵ + k̂, \(\vec{b}\) = î + 0 ∙ ĵ + 0 ∙ k̂ and \(\vec{c}\) = c_{1}î + c_{2}ĵ + c_{3}k̂, then

(a) Let c_{1} = 1 and c_{2} = 2, find c_{3} which makes \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) coplanar.

(b) If c_{2} = – 1 and c_{3} = 1, show that no value of c_{1} can make \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) coplanar. (Delhi 2017)

Answer:

Given, \(\vec{a}\) = î + ĵ + k̂, \(\vec{b}\) = î + 0 ∙ ĵ + 0 ∙ k̂ and \(\vec{c}\) = c_{1}î + c_{2}ĵ + c_{3}k̂

The given vectors are coplanar iff \([\vec{a} \vec{b} \vec{c}]\) = 0

(a) If c_{1} = 1 and c_{2} = 2,

Then, from Eq.(i), we get

\(\left|\begin{array}{ccc}

1 & 1 & 1 \\

1 & 0 & 0 \\

1 & 2 & c_{3}

\end{array}\right|\) = 0

⇒ – 1(c_{3} – 0) + 1(2 – 0) = 0

⇒ – c_{3} + 2 = 0

⇒ – c_{3} = – 2

⇒ c_{3} = 2

(b) If c_{2} = – 1 and c_{3} = 1, then from Eq. (i), we get

\(\left|\begin{array}{ccc}

1 & 1 & 1 \\

1 & 0 & 0 \\

c_{1} & -1 & 1

\end{array}\right|\) = 0

⇒ 1(0) – 1(1 – 0) + 1(- 1 – 0) = 0

⇒ 0 – 1 – 1 = 0

⇒ – 2 ≠ 0

∴ No value of c1 can make 1’ and coplanar.

Hence proved

Question 54.

Show that the points A, B, C with position vectors 2î – ĵ + k̂, î – 5ĵ – 5k̂ and 5î – 4ĵ – 4k̂ respectively, are the vertices of a right-angled triangle. Hence find the area of the triangle. (All India 2017)

Answer:

Question 55.

Show that the vectors \(\vec{a}\), \(\vec{b}\) and \(\vec{c}\) are coplanar, if a + b, 6+ c and c+ a are coplanar. (Delhi 2016, Foreign 2014)

Or

Prove that, for any three vectors \(\vec{a}\), \(\vec{b}\) and \(\vec{c},[\vec{a}+\vec{b} \vec{b}+\vec{c} \vec{c}+\vec{a}]=2[\vec{a} \vec{b} \vec{c}]\). (Delhi 2014)

Answer:

Question 56.

Show that the four points A (4, 5, 1), B(0, -1, -1), C(3, 9, 4) and D (-4, 4, 4) are coplanar. (All India 2016)

Or

Show that the four points A, B, C and D with position vectors 4î + 5ĵ + k̂, – ĵ – k̂, 3î + 9ĵ + 4k̂ and 4(- î + ĵ + k̂), respectively are coplanar. (All India 2014)

Answer:

Let the position vector of points A, B, C and D are

= – 4(12 + 3) + 6 (- 3 + 24) – 2(1 + 32)

= – 60 + 126 – 66 = 0

Hence, the four points A, B, C and D are coplanar.

Question 57.

The two adjacent sides of a parallelogram are 2î – 4ĵ – 5k̂ and 2î + 2 ĵ + 3k̂. Find the two unit vectors parallel to its diagonals. Using the diagonal vectors, find the area of the parallelogram. (All India 2016)

Answer:

Let ABCD be the given parallelogram with

Question 58.

If \(\vec{a} \times \vec{b}=\vec{c} \times \vec{d}\) and \(\vec{a} \times \vec{c}=\vec{b} \times \vec{d}\), then show that \(\vec{a}-\vec{d}\) is parallel to \(\vec{b}-\vec{c}\), where \(\vec{a} \neq \vec{d}\) and \(\vec{b} \neq \vec{c}\). (Foreign 2016; Delhi 2009)

Answer:

Use the result, if two vectors are parallel, then their cross-product will be a zero vector.

Question 59.

If \(\vec{r}\) = xî + yĵ + zk̂, find \((\vec{r} \times \hat{i}) \cdot(\vec{r} \times \hat{j})\) + xy. (Delhi 2015)

Answer:

Question 60.

If \(\vec{a}\) = î + 2ĵ + k̂, \(\vec{b}\) = 2î + ĵ and \(\vec{c}\) = 3î – 4 ĵ – 5k̂, then find a unit vector perpendicular to both of the vectors \((\vec{a}-\vec{b})\) and \((\vec{c}-\vec{b})\). (All India 2015)

Answer:

Question 61.

Find the value of λ so that the four points A, B,C and D with position vectors 4 î + 5ĵ + k̂, -ĵ – k̂,3i + Xj+4k and – 4 î + 4ĵ + 4 k̂, respectively are coplanar. (Delhi 2015C)

Answer:

Use the condition that four points with position vectors \(\vec{A}, \vec{B}, \vec{C}\) and \(\vec{D}\) are coplanar, if

\([\overrightarrow{A B}, \overrightarrow{A C}, \overrightarrow{A D}]=\overrightarrow{0}\) = 0.

On expanding along R_{1}, we get

⇒ – 4(3λ – 15 + 3) + 6(- 3 + 24) – 2(1 + 8λ – 40) = 0

⇒ – 4(3λ – 12) + 6(21) – 2(8λ – 39) = 0

⇒ – 12λ + 48 + 126 – 16λ + 78 = 0

⇒ – 28λ + 252 = 0

λ = 9

Question 62.

Prove that \(\left[\begin{array}{lll}

\vec{a} & \vec{b}+\vec{c} & \vec{d}

\end{array}\right]=\left[\begin{array}{lll}

\vec{a} & \vec{b} & \vec{d}

\end{array}\right]+\left[\begin{array}{lll}

\vec{a} & \vec{c} & \vec{d}

\end{array}\right]\). (All India 2015C)

Answer:

Question 63.

If \(\vec{a}\) = 2î – 3ĵ + k̂, \(\vec{b}\) = – î + k̂, \(\vec{c}\) = 2 ĵ – k̂ are three vectors, find the area of the parallelogram having diagonals \(\) and \(\). (Delhi 2014C)

Answer:

Question 64.

Vectors \(\vec{a}, \vec{b}\) and \(\vec{c}\) are such that \(\vec{a}+\vec{b}+\vec{c}=\overrightarrow{0}\) and |\(\vec{a}\)| = 3, |\(\vec{b}\)| = 5 and |\(\vec{c}\)| = 7. Find the angle between \(\vec{a}\) and \(\vec{b}\). (Delhi 2014,2008; All India 2008)

Answer:

\(\frac{\pi}{3}\)

Question 65.

The scalar product of the vector \(\vec{a}\) = î + ĵ + k̂ with a unit vector along the sum of vectors \(\vec{b}\) = 2î + 4ĵ – 5k̂ and \(\vec{c}\) = λî + 2ĵ + 3k̂ is equal to one. Find the value of λ and hence, find the unit vector along \(\vec{b}\) + \(\vec{c}\). (All India 2014)

Or

The scalar product of vector i + j + k with the unit vector along the sum of vectors 2î + 4ĵ – 5k̂ and λî + 2ĵ + 3k̂ is equal to one. Find the value of λ. (All India 2009,2008C)

Answer:

First, determine the unit vector of \(\vec{b}+\vec{c}\), i.e. \(\frac{\vec{b}+\vec{c}}{|\vec{b}+\vec{c}|}\). Further put \(\vec{a} \cdot \frac{(\vec{b}+\vec{c})}{|\vec{b}+\vec{c}|}\) = 1 and then determine the value of λ.

⇒ (λ + 6)^{2} = λ^{2} + 4λ + 44 [squaring both sides]

⇒ λ^{2} + 36 + 12λ + 4λ + 44

⇒ 8λ = 8

⇒ λ = 1

Hence, the value of λ is 1.

On substituting the value of λ in Eq. (1), we get Unit vector along \(\vec{b}+\vec{c}\)

Question 66.

Find the vector \(\vec{p}\) which is perpendicular to both \(\vec{α}\) = 4î + 5ĵ – k̂ and \(\vec{β}\) = î – 4ĵ + 5k̂ and \(\vec{p}\). \(\vec{q}\) = 21, where \(\vec{q}\) = 3i + j – k. (All India 2014C)

Answer:

= î (25 – 4) – ĵ (20 + 1) + k̂(- 16 – 5)

= î(21) – ĵ(21) + k̂(- 21)

= 21î – 21ĵ – 21k̂

So, \(\vec{p}\) = 21λk̂—21λ?—21λk [fromEq.(i)] ….. (ii)

Also, given that \(\vec{p} \cdot \vec{q}\) = 21

∴ (21λî – 21λĵ – 21λk̂) . (3î + ĵ – k̂) = 21

⇒ 63λ – 21λ + 21λ = 21

⇒ 63λ = 21

⇒ λ = \(\frac{1}{3}\)

On putting λ = \(\frac{1}{3}\) in Eq. (ii), we get

which is the required vector.

Question 67.

Find the unit vector perpendicular to both of the vectors \(\vec{a}+\vec{b}\) and \(\vec{a}-\vec{b}\) where, \(\vec{a}\) = î + ĵ + k̂ and \(\vec{b}\) = î + 2ĵ + 3k̂. (Foreign 2014)

Answer:

⇒ 2x + 3y + 4z = 0 …… (ii)

and (xî + yĵ + zk̂) . (- ĵ – 2k̂) = 0

⇒ – y – 2z = 0

⇒ y = – 2z

On putting the value of yin Eq. (ii), we get

2x + 3 (- 2z) + 4z = 0

⇒ x = z

On substituting the value of x and y in Eq. (1),

we get

⇒ z^{2} + 4z^{2} + z^{2} = 1

⇒ 6z^{2} = 1

⇒ z = ± \(\frac{1}{\sqrt{6}}\)

then, x = ± \(\frac{1}{\sqrt{6}}\)

and y = ± \(\frac{2}{\sqrt{6}}\)

Hence, the required vectors are

Question 68.

Find the unit vector perpendicular to the plane ABC where the position vectors of A, B and C are 2î – ĵ + k̂, î + ĵ + 2k̂ and 2î + 3k̂, respectively. (All India 2014C)

Answer:

A unit vector perpendicular to plane ABC is

\(\frac{\overrightarrow{A B} \times \overrightarrow{A C}}{|\overrightarrow{A B} \times \overrightarrow{A C}|}\)

Let O be the origin of reference.

Question 69.

Dot product of a vector with vectors î – ĵ + k̂, 2î + ĵ – 3k̂ and î + ĵ + k̂ are respectively 4, 0 and 2. Find the vector. (Delhi 2013C)

Answer:

⇒ a_{1} + a_{2} + a_{3} = 2

On subtracting Eq. (iii) from Eq. (i), we get

– 2a_{2} = 2

⇒ a_{2} = – 1

On substituting a_{2} = – 1 in Eq. (ii) and (iii),

we get

2a_{2} – 3a_{3} = 1 …… (iv)

⇒ a_{1} + a_{3} = 3

On multiplying Eq. (v) by 3 and then adding with Eq. (iv), we get

5a_{1} = 1 + 9 = 10

⇒ a_{1} = 2

On substituting a_{1} = 2 in Eq. (v), we get

a_{3} = 1

Hence, the vector is \(\vec{a}\) = 2î – ĵ + k̂

Question 70.

Find the values of λ for which the angle between the vectors \(\vec{a}\) = 2λ^{2}î + 4λĵ + k̂ and \(\vec{b}\) = 7î – 2ĵ + λk̂ is obtuse. (All India 2013C)

Answer:

let θ be the obtuse angle between the vectors

14λ^{2} – 7λ < 0

= 2λ^{2} – λ < 0

Either λ < 0, 2λ – 1 > 0 or λ > 0, 2λ – 1 < 0

= Either λ < 0, λ > \(\frac{1}{2}\) or λ > 0, λ < \(\frac{1}{2}\) Clearly, first option is impossible. ∴ λ > 0, λ < \(\frac{1}{2}\)

0 < λ < \(\frac{1}{2}\)

λ ∈ \(\left(0, \frac{1}{2}\right)\)

Question 71.

If a, b and c are three vectors such that each one is perpendicular to the vector obtained by sum of the other two and |\(\vec{a}\)| = 3, |\(\vec{b}\)| = 4 and |\(\vec{c}\)| = 5, then prove that |\(\vec{a}+\vec{b}+\vec{c}\)| = 5√2. (All India 2013C, 2010C)

Or

If \(\vec{a}, \vec{b}\) and \(\vec{c}\) are three vectors, such that |\(\vec{a}\)| = 3, |\(\vec{b}\)| = 4 and |\(\vec{c}\)| = 5 and each one of these is perpendicular to the sum of other two, then find |\(\vec{a}+\vec{b}+\vec{c}\)|. (All India 2011C, 2010C)

Answer:

Question 72.

If \(\vec{a}\) = 3î – ĵ and \(\vec{b}\) = 2î + ĵ – 3k̂, then express \(\vec{b}\) in the form \(\vec{b}=\vec{b}_{1}+\vec{b}_{2}\), where \(\vec{b}_{1} \| \vec{a}\) and \(\vec{b}_{2} \perp \vec{a}\). (All India 2013C)

Answer:

Given \(\vec{a}\) = 3î – ĵ and \(\vec{b}\) = 2î + ĵ – 3k̂

Let \(\overrightarrow{b_{1}}\) = x_{1}î + y_{1}ĵ + z_{1}k̂ are two vectors such that \(\overrightarrow{b_{1}}+\overrightarrow{b_{2}}=\vec{b}, \overrightarrow{b_{1}} \| \vec{a}\) and \(\overrightarrow{b_{2}} \perp \vec{a}\)

Consider, \(\vec{b}_{1}+\vec{b}_{2}=\vec{b}\)

⇒ (x_{1} + x_{2})î + (y_{1} + y_{2})ĵ + (z_{1} + z_{2})k̂ = 2î + ĵ – 3k̂

On comparing the coefficient of î ĵ and k̂ both sides, we get

x_{1} + x_{2} = 2

y_{1} + y_{2} = 1

z_{1} + z_{2} = -3

Now, consider \(\overrightarrow{b_{1}} \| \vec{a}\)

⇒ \(\frac{x_{1}}{3}=\frac{y_{1}}{-1}=\frac{z_{1}}{0}\)

⇒ x_{1} = 3λ, y_{1} = -λ,and z_{1} = 0 …(iv)

On substituting the values of x, y and z, from Eq. (iv) to Eq. (i), (ii) and (iii), respectively, we get

x_{2} = 2- 3λ, y_{2} = -1 + λ and z_{2} = -3 …(v)

Since, b2 ± a , therefore b2 a = 0

⇒ 3x_{2} – y_{2} = 0

⇒ 3 (2 – 3λ) – (1 + λ) = 0

⇒ 6 – 9λ – 1 – λ = 0

⇒ 5 – 10λ = 0

⇒ λ = \(\frac{1}{2}\)

On substituting λ = \(\frac{1}{2}\) in Eqs. (iv) and (v), we get

Question 73.

If \(\vec{a}\) = î + ĵ + k̂ and \(\vec{b}\) = ĵ – k̂, then find a vector \(\vec{c}\), such that \(\vec{a} \times \vec{c}=\vec{b}\) and \(\vec{a} \cdot \vec{c}\) = 3. (Delhi 2013, 2008)

Answer:

Given \(\vec{a}\) = î + ĵ + k̂ and \(\vec{b}\) = ĵ – k̂

Let \(\vec{c}\) = xî + yĵ + zk̂

= î (z – y) – ĵ(z – x) + k̂(y – x)

Now, \(\vec{a} \times \vec{c}=\vec{b}\) [given]

= î(z – y) + ĵ(x – z) + k̂(y – x)

= 0î + 1ĵ + (-1)k̂ [∵ \(\vec{b}\) = ĵ – k̂]

On comparing the coefficients from both sides, we get

z – y = 0,x – z = 1, y – x = -1

⇒ y = z and x – y = 1…(i)

Also given, \(\vec{a} \cdot \vec{c}\) = 3

⇒ (î + ĵ + k̂) . (xî + yĵ + zk̂) = 3

⇒ x + y + z = 3 (1)

⇒ x + 2y = 3 [∵ y = z] …(ii)

On subtracting Eq. (i) from Eq. (ii), we get

3y = 2

⇒ y = \(\frac{2}{3}\) = z [∵ y = z]

From Eq. (i),

x = 1 + y + 1 = 1 + \(\frac{2}{3}=\frac{5}{3}\)

Hence, \(\vec{c}=\frac{5}{3} \hat{i}+\frac{2}{3} \hat{j}+\frac{2}{3} \hat{k}\)

Question 74.

If \(\vec{a}\) = î – ĵ + 7k̂ and \(\vec{b}\) = 5î – ĵ + λk̂, then find the value of λ, so that \(\) and \(\) are perpendicular vectors. (All India 2013)

Answer:

Use the result that if \(\vec{a}\) and \(\vec{b}\) are perpendicular, then their dot product should be zero and simplify it.

Given, \(\vec{a}\) = î – ĵ + 7k̂ and \(\vec{b}\) = 5î – ĵ + λk̂

Then, \(\vec{a}+\vec{b}\) = (î – ĵ + 7k̂) + (5î – ĵ + λk̂)

= 6î – 2ĵ + (7 + λ) k̂

and a – = (î – ĵ + 7k̂) – (5î – j ̂+ λk̂)

= -4î + (7 – λ)k̂

Since, \((\vec{a}+\vec{b})\) and \((\vec{a}-\vec{b})\) are perpendicular

vectors, then \((\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})\) = 0

⇒ [6î – 2ĵ + (7 + λ)k̂]- [-4î + (7 – λ)k̂] = 0 (1)

⇒ -24 + (7+ X)(7 – X) =0

⇒ 49 – λ^{2} = 24

⇒ λ^{2} = 25

∴ λ = ± 5

Question 75.

If p = 5î + λĵ – 3k̂ and q = î + 3ĵ – 5k̂, then find the value of λ, so that \(\vec{p}+\vec{q}\) and \(\vec{p}-\vec{q}\) are perpendicular vectors. (All India 2013)

Answer:

λ = ± 1

Question 76.

If \(\vec{a}, \vec{b}\) and \(\vec{c}\) are three vectors, such that |\(\vec{a}\)| = 5, |\(\vec{b}\)| = 12, |\(\vec{c}\)| = 13 and \(\vec{a}+\vec{b}+\vec{c}\) = 0, then find the value of \(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}\). (Delhi 2012)

Answer:

-169

Question 77.

Let \(\vec{a}\) = î + 4ĵ + 2k̂, \(\vec{b}\) = 3î – 2ĵ + 7k̂ and \(\vec{c}\) = 2î – ĵ + 4k̂. Find a vector \(\vec{p}\), which is perpendicular to both \(\vec{a}\) and \(\vec{b}\) and \(\vec{p}.\vec{c}\) = 18. (All India 2012,2010)

Answer:

Given vectors are \(\vec{a}\) = î + 4ĵ + 2k̂,

\(\vec{b}\) = 3î – 2ĵ + 7k̂

and \(\vec{c}\) = 2î – ĵ + 4k̂

Let \(\vec{p}\) = xî + yĵ + zk̂

We have, \(\vec{p}\) is perpendicular to both \(\vec{a}\) and \(\vec{b}\).

\(\vec{p} \cdot \vec{a}\) = 0

⇒ (xî + yĵ + zk̂) – (î + 4ĵ + 7k̂) = 0

⇒ x + 4y + 2z = 0 ………….(i)

and \(\vec{p} \cdot \vec{b}\) = 0

⇒ (xî + yĵ + zk̂) . (3î – 2ĵ + 7k̂) = 0

⇒ 3x – 2y + 7z = 0 …(ii)

Also, given ~p-~c =18 (1)

⇒ (xî + yĵ + zk̂) . (2î – ĵ + 4k̂) = 0

⇒ 2x – y + 4z = 18 …(iii)

On multiplying Eq. (i) by 3 and subtracting it from Eq. (ii), we get

– 14y + z = 0 ..(iv)

Now, multiplying Eq. (i) by 2 and subtracting it from Eq. (iii), we get

– 9y = 18

⇒ y = -2

On putting y = -2 in Eq. (iv), we get

-14 (-2) + z = 0

⇒ 28 + z = 0

⇒ z = -28

On putting y = -2 and z = -28 in Eq. (i), we get

x + 4 (-2) + 2 (-28) = 0

⇒ x – 8 – 56 = 0

⇒ x = 64

Hence, the required vector is

\(\vec{p}\) = xî + yĵ + zk̂

i.e. \(\vec{p}\) = 64î – 2ĵ – 28k̂

Question 78.

Find a unit vector perpendicular to each of the vectors \(\vec{a}+\vec{b}\) and \(\vec{a}-\vec{b}\), where a = 3î + 2 ĵ + 2k̂ and b = î + 2ĵ – 2k̂. (Delhi 2011)

Answer:

\(\frac{2}{3} \hat{i}-\frac{2}{3} \hat{j}-\frac{1}{3} \hat{k}\)

Question 79.

If a and 6 are two vectors, such that |\(\vec{a}\)| = 2, |\(\vec{b}\)| = 1 and \(\vec{a}\).\(\vec{b}\) = 1, then find \((3 \vec{a}-5 \vec{b}) \cdot(2 \vec{a}+7 \vec{b})\). (Delhi 2011)

Answer:

Question 80.

If vectors \(\vec{a}\) = 2î + 2ĵ + 3k̂, \(\vec{b}\) = -î + 2ĵ + k̂ and \(\vec{c}\) = 3î + ĵ are such that \(\vec{a}+\lambda \vec{b}\) is perpendicular to \(\vec{c}\), then find the value λ. (Foreign 2011; All India 2009C)

Answer:

Given, \(\vec{a}\) = 2î + 2ĵ + 3k̂,

\(\vec{b}\) = -î + 2ĵ + k̂

and \(\vec{c}\) = 3î + ĵ

Also, \(\vec{a}\) + λ\(\vec{b}\) is perpendicular to \(\vec{c}\).

∴ (\(\vec{a}\) + λ\(\vec{b}\)).\(\vec{c}\) = 0 …(i) [∵ when \([latex]\)[/latex], then \(\vec{a} \cdot \vec{b}\) = 0]

Now, \(\vec{a}\) + λ\(\vec{b}\) = (2î + 2ĵ + 3k̂) + λ (-î + 2ĵ + k̂)

⇒ \(\vec{a}\) + λ\(\vec{b}\) = î(2 – λ) + ĵ(2 + 2λ) + k̂(3 + λ)

Then, from Bq. (i), we get

[î (2 – λ) + ĵ (2 + 2λ) + k̂(3 + λ)].[3î + ĵ] = 0

⇒ 3(2 – λ) + 1(2+ 2k) = 0

⇒ 8 – λ = 0

∴ λ = 8

Question 81.

Using vectors, find the area of triangle with vertices A (1, 1, 2), 5(2, 3, 5) and C (1, 5, 5). (All India 2011)

Answer:

\(\frac{1}{2}\)\(\sqrt{61}\) sq.units

Question 82.

Using vectors, find the area of triangle with vertices A (2, 3, 5), B (3, 5, 8) and C(2, 7, 8). (Delhi 2010C)

Answer:

\(\frac{1}{2}\)\(\sqrt{61}\) sq.units