If a → = 3 i ^ + 2 j ^ + 9 k ^ and b → = i ^ + λ j ^ + 3 k ^ , find the value of λ so that a → + b → is perpendicular to a → - b → .

The first step is to find the sum and difference of the two vectors, a → + b → = 3 i ^ + 2 j ^ + 9 k ^ + i ^ + λ j ^ + 3 k ^ = 4 i ^ + 2 + λ j ^ + 12 k ^ a → - b → = 3 i ^ + 2 j ^ + 9 k ^ - i ^ + λ j ^ + 3 k ^ = 2 i ^ + 2 - λ j ^ + 6 k ^ Since the above two vectors are perpendicular, their scalar product is 0 . a → + b → · a → - b → = 0 ⇒ 4 i ^ + 2 + λ j ^ + 12 k ^ · 2 i ^ + 2 - λ j ^ + 6 k ^ = 0 ⇒ 4 × 2 + 2 + λ 2 - λ + 12 × 6 = 0 ⇒ 8 + 2 2 - λ 2 + 72 = 0 ⇒ - λ 2 = - 84 ⇒ λ 2 = 84 ⇒ λ = 84 = ± 2 21 .

If a → = 4 i ^ + 2 j ^ - k ^ and b → = 5 i ^ + 2 j ^ - 3 k ^ , find the angle between the vectors a → + b → and a → - b →

We need to compute the vectors, a → + b → = 4 i ^ + 2 j ^ - k ^ + 5 i ^ + 2 j ^ - 3 k ^ = 9 i ^ + 4 j ^ - 4 k ^ a → - b → = 4 i ^ + 2 j ^ - k ^ - 5 i ^ + 2 j ^ - 3 k ^ = - i ^ + 2 k ^ The formula to find the angle between two vectors is given by cos θ = a → + b → · a → - b → a → + b → a → - b → . Finding the scalar product, a → + b → · a → - b → = 9 i ^ + 4 j ^ - 4 k ^ · - i ^ + 2 k ^ = 9 × - 1 + 4 × 0 + - 4 × 2 = - 9 + 0 - 8 = - 17 Finding the magnitude of vectors, a → + b → = 9 2 + 4 2 + - 4 2 = 81 + 16 + 16 = 113 a → - b → = - 1 2 + 2 2 = 1 + 4 = 5 Therefore, the angle between the vectors is cos θ = - 17 113 × 5 = - 17 565 ⇒ θ = cos - 1 - 17 565 = π - cos - 1 17 565 .

For what value of λ , the vectors a → = 2 i ^ + λ j ^ + k ^ and b → = i ^ - 2 j ^ + 3 k ^ perpendicular to each other?

In order to get the values of λ for which the vectors a → = 2 i ^ + λ j ^ + k ^ and b → = i ^ - 2 j ^ + 3 k ^ are perpendicular, the condition is that their dot product must be zero. a → · b → = 0 ⇒ 2 i ^ + λ j ^ + k ^ · i ^ - 2 j ^ + 3 k ^ = 0 ⇒ 2 × 1 + λ × - 2 + 1 × 3 = 0 ⇒ 5 - 2 λ = 0 λ = 5 2 .

If a → = 2 i ^ - j ^ + k ^ , b → = i ^ - 3 j ^ - 5 k ^ , then find a vector c → such that a → , b → , c → form the sides of a right-angled triangle taken in order.

Here, a → · b → = 2 + 3 - 5 = 0 Hence, a → & b → are perpendicular. Since the two sides of a triangle are given, using triangle law of vector addition, the third side is given by their sum c → = a → + b → = 2 i ^ - j ^ + k ^ + i ^ - 3 j ^ - 5 k ^ = 3 i ^ - 4 j ^ - 4 k ^

Find the vector of magnitude 3 2 which lies in the z x -plane and is at right angles to the vector 2 i ^ + j ^ + 2 k ^ .

Since the vector lies in the z x -plane, let's consider the vector to be found as x i ^ + z k ^ . The magnitude of the vector is given as 3 2 , so x i ^ + z k ^ = x 2 + z 2 = 3 2 ⇒ x 2 + z 2 = 9 × 2 ⇒ x 2 + z 2 = 18 ⋯ 1 Also, it is said that the vector is at right angles to the vector 2 i ^ + j ^ + 2 k ^ . Their scalar product must be zero. x i ^ + z k ^ · 2 i ^ + j ^ + 2 k ^ = 0 ⇒ 2 x + 2 z = 0 ⇒ x = - z Substituting x = - z in Equation ( 1 ) , - z 2 + z 2 = 18 ⇒ z 2 + z 2 = 18 ⇒ 2 z 2 = 18 ⇒ z 2 = 9 ⇒ z = ± 3 When z = 3 , x = - 3 and when z = - 3 , x = 3 . So the required vector can be - 3 i ^ + 3 k ^ or 3 i ^ - 3 k ^ . It can be rewritten as ± 3 i ^ - k ^ .

Find the values of x for which the angle between the vectors a → = - 3 i ^ + x j ^ + k ^ and b → = x i ^ + 2 x j ^ + k ^ is acute and the angle between b → and x -axis lies between π 2 and π .

It is said in the question that the angle between a → = - 3 i ^ + x j ^ + k ^ and b → = x i ^ + 2 x j ^ + k ^ is acute. So, a → · b → > 0 . - 3 i ^ + x j ^ + k ^ · x i ^ + 2 x j ^ + k ^ > 0 ⇒ - 3 x + 2 x 2 + 1 > 0 ⇒ 2 x 2 - 3 x + 1 > 0 ⇒ 2 x 2 - 2 x - x + 1 > 0 ⇒ 2 x x - 1 - x - 1 > 0 ⇒ 2 x - 1 x - 1 > 0 x < 1 2 or x > 1 … 1 (by using wavy curve method) It is also said in the question that the angle between b → and x -axis lies between π 2 and π , which means that the angle is obtuse. Consider a vector c → = i ^ that lies on the x -axis. So, b → · c → < 0 . ( x i ^ + 2 x j ^ + k ^ ) · i ^ < 0 x < 0 . . . ( 2 ) From 1 and 2 x < 0 .

Let a → = i ^ + 4 j ^ + 2 k ^ , b → = 3 i ^ - 2 j ^ + 7 k ^ and c → = 2 i ^ - j ^ + 4 k ^ . Find a vector d → which is perpendicular to both a → and b → and satisfies c → · d → = 15 .

Consider d → = x i ^ + y j ^ + z k ^ . Now since the vector is perpendicular to both a → and b → ⇒ d → · a → = d → · b → = 0 d → · a → ⇒ x i ^ + y j ^ + z k ^ · i ^ + 4 j ^ + 2 k ^ = 0 ⇒ x + 4 y + 2 z = 0 … 1 d → · b → ⇒ x i ^ + y j ^ + z k ^ · 3 i ^ - 2 j ^ + 7 k ^ = 0 ⇒ 3 x - 2 y + 7 z = 0 … 2 Also, c → · d → = 15 , so 2 i ^ - j ^ + 4 k ^ · x i ^ + y j ^ + z k ^ = 15 ⇒ 2 x - y + 4 z = 15 … 3 We need to solve the set of Equations 1 , 2 and 3 . From Equation 3 , we can write y in terms of x and z and then substitute that in other equations. 2 x - y + 4 z = 15 ⇒ y = 2 x + 4 z - 15 Equation ( 1 ) x + 4 2 x + 4 z - 15 + 2 z = 0 ⇒ 9 x + 18 z - 60 = 0 ⇒ 3 x + 6 z - 20 = 0 ⇒ 3 x + 6 z = 20 … 4 Equation ( 2 ) 3 x - 2 ( 2 x + 4 z - 15 ) + 7 z = 0 ⇒ - x - z + 30 = 0 x + z = 30 … 5 Multiplying Equation 5 with 3 and solving Equations 4 and 5 3 x + 6 z = 20 - 3 x + 3 z = 90 3 z = - 70 z = - 70 3 Substituting this in Equation ( 5 ) , x = 30 - - 70 3 = 90 + 70 3 = 160 3 Substituting values of x and z in equation for y y = 2 160 3 + 4 - 70 3 - 15 = 320 3 - 280 3 - 45 3 = - 5 3 Therefore, d → = 160 3 i ^ - 5 3 j ^ - 70 3 k ^ .

E M B I B E

VECTOR AND 3-D GEOMETRY for JEE Main and Advanced>Vectors>Chapter 2 - Scalar or Dot Product of Two Vectors>EXERCISE

K. C. Sinha and Abhishek Chandra Solutions for Exercise 1: EXERCISE

Author:K. C. Sinha & Abhishek Chandra

K. C. Sinha Mathematics Solutions for Exercise - K. C. Sinha and Abhishek Chandra Solutions for Exercise 1: EXERCISE

Attempt the practice questions from Exercise 1: EXERCISE with hints and solutions to strengthen your understanding. VECTOR AND 3-D GEOMETRY for JEE Main and Advanced solutions are prepared by Experienced Embibe Experts.

Questions from K. C. Sinha and Abhishek Chandra Solutions for Exercise 1: EXERCISE with Hints & Solutions

MEDIUM

JEE Advanced

IMPORTANT

VECTOR AND 3-D GEOMETRY for JEE Main and Advanced>Chapter 2 - Scalar or Dot Product of Two Vectors>EXERCISE>Q 9

If $\vec{a} = 3 \hat{i} + 2 \hat{j} + 9 \hat{k}$ and $\vec{b} = \hat{i} + λ \hat{j} + 3 \hat{k}$ , find the value of $λ$ so that $\vec{a} + \vec{b}$ is perpendicular to $\vec{a} - \vec{b}$ .

MEDIUM

JEE Advanced

IMPORTANT

VECTOR AND 3-D GEOMETRY for JEE Main and Advanced>Chapter 2 - Scalar or Dot Product of Two Vectors>EXERCISE>Q 10

If $\vec{a} = 4 \hat{i} + 2 \hat{j} - \hat{k}$ and $\vec{b} = 5 \hat{i} + 2 \hat{j} - 3 \hat{k}$ , find the angle between the vectors $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$

MEDIUM

JEE Advanced

IMPORTANT

VECTOR AND 3-D GEOMETRY for JEE Main and Advanced>Chapter 2 - Scalar or Dot Product of Two Vectors>EXERCISE>Q 12

For what value of $λ$ , the vectors $\vec{a} = 2 \hat{i} + λ \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} - 2 \hat{j} + 3 \hat{k}$ perpendicular to each other?

HARD

JEE Advanced

IMPORTANT

VECTOR AND 3-D GEOMETRY for JEE Main and Advanced>Chapter 2 - Scalar or Dot Product of Two Vectors>EXERCISE>Q 17

Find the scalar components of a unit vector which is perpendicular to each of the vectors $\hat{i} + 2 \hat{j} - \hat{k}$ and $3 \hat{i} - \hat{j} + 2 \hat{k}$ .

MEDIUM

JEE Advanced

IMPORTANT

VECTOR AND 3-D GEOMETRY for JEE Main and Advanced>Chapter 2 - Scalar or Dot Product of Two Vectors>EXERCISE>Q 18

If $\vec{a} = 2 \hat{i} - \hat{j} + \hat{k}, \vec{b} = \hat{i} - 3 \hat{j} - 5 \hat{k}$ , then find a vector $\vec{c}$ such that $\vec{a}, \vec{b}, \vec{c}$ form the sides of a right-angled triangle taken in order.

HARD

JEE Advanced

IMPORTANT

VECTOR AND 3-D GEOMETRY for JEE Main and Advanced>Chapter 2 - Scalar or Dot Product of Two Vectors>EXERCISE>Q 19

Find the vector of magnitude $3 \sqrt{2}$ which lies in the $z x$ -plane and is at right angles to the vector $2 \hat{i} + \hat{j} + 2 \hat{k}$ .

HARD

JEE Advanced

IMPORTANT

VECTOR AND 3-D GEOMETRY for JEE Main and Advanced>Chapter 2 - Scalar or Dot Product of Two Vectors>EXERCISE>Q 20

Find the values of $x$ for which the angle between the vectors $\vec{a} = - 3 \hat{i} + x \hat{j} + \hat{k}$ and $\vec{b} = x \hat{i} + 2 x \hat{j} + \hat{k}$ is acute and the angle between $\vec{b}$ and $x$ -axis lies between $\frac{π}{2}$ and $π$ .

HARD

JEE Advanced

IMPORTANT

VECTOR AND 3-D GEOMETRY for JEE Main and Advanced>Chapter 2 - Scalar or Dot Product of Two Vectors>EXERCISE>Q 22

Let $\vec{a} = \hat{i} + 4 \hat{j} + 2 \hat{k}, \vec{b} = 3 \hat{i} - 2 \hat{j} + 7 \hat{k}$ and $\vec{c} = 2 \hat{i} - \hat{j} + 4 \hat{k}$ . Find a vector $\vec{d}$ which is perpendicular to both $\vec{a}$ and $\vec{b}$ and satisfies $\vec{c} \cdot \vec{d} = 15$ .

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