If a → = 2 i ^ + 3 j ^ - 4 k ^ , b → = 3 i ^ - 4 j ^ - 5 k ^ and c → = - 3 i ^ + 2 j ^ + 3 k ^ , find the magnitude and direction cosines of a → + b → + c →

Given that, the vectors are a → = 2 ı ^ + 3 ȷ ^ - 4 k ^ b → = 3 ı ^ - 4 ȷ ^ - 5 k ^ c → = - 3 ı ^ + 2 ȷ ^ + 3 k ^ Consider, a → + b → + c → = ( 2 i ^ + 3 j ^ - 4 k ^ ) + ( 3 i ^ - 4 j ^ - 5 k ^ ) + ( - 3 i ^ + 2 j ^ + 3 k ^ ) a → + b → + c → = 2 i ^ + j ^ - 6 k ^ | a → + b → + c → | = | 2 i ^ + j ^ - 6 k ^ | = 2 2 + 1 2 + ( - 6 ) 2 = 41 Direction cosines of 2 i ^ + j ^ - 6 k ^ are 2 | 2 i ^ + j ^ - 6 k ^ | , 1 | 2 i ^ + j ^ - 6 k ^ | , - 6 | i ^ + j ^ - 6 k ^ | 2 41 , 1 41 , - 6 41 ∴ The magnitude and direction cosines of the vector a → + b → + c → are 41 , 2 41 , 1 41 , - 6 41 .

If a → = 2 i ^ + 3 j ^ - 4 k ^ , b → = 3 i ^ - 4 j ^ - 5 k ^ and c → = - 3 i ^ + 2 j ^ + 3 k ^ , find the magnitude and direction cosines of 3 a → - 2 b → + 5 c →

Given that, the vectors are a → = 2 ı ^ + 3 ȷ ^ - 4 k ^ b → = 3 ı ^ - 4 ȷ ^ - 5 k ^ c → = - 3 ı ^ + 2 ȷ ^ + 3 k ^ Consider, 3 a → - 2 b → + 5 c → = 3 ( 2 i ^ + 3 j ^ - 4 k ^ ) - 2 ( 3 i ^ - 4 j ^ - 5 k ^ ) + 5 ( - 3 i ^ + 2 j ^ + 3 k ^ ) = 6 i ^ + 9 j ^ - 12 k ^ - 6 i ^ + 8 j ^ + 10 k ^ - 15 i ^ + 10 j ^ + 15 k ^ ∴ 3 a → - 2 b → + 5 c → = - 15 i ^ + 27 j ^ + 13 k ^ Now, | 3 a → - 2 b → + 5 c → | = | - 15 i ^ + 27 j ^ + 13 k ^ | = ( - 15 ) 2 + ( 27 ) 2 + 13 2 | 3 a → - 2 b → + 5 c → | = 1123 Direction cosines of the vector 3 a → - 2 b → + 5 c → are - 15 | - 15 i ^ + 27 j ^ + 13 k ^ | , 27 | - 15 i ^ + 27 j ^ + 13 k ^ | , 13 | - 15 i ^ + 27 j ^ + 13 k ^ | = - 15 1123 , 27 1123 , 13 1123 ∴ The magnitude and direction cosines of the vector 3 a → - 2 b → + 5 c → are 1123 , - 15 1123 , 27 1123 , 13 1123 .

The position vectors of the vertices of a triangle are i ^ + 2 j ^ + 3 k ^ , 3 i ^ - 4 j ^ + 5 k ^ and - 2 i ^ + 3 j ^ - 7 k ^ . Find the perimeter of the triangle.

Let A , B , C be the vertices of a triangle ABC . Given that, the position vectors of the vertices A , B , C are OA → = i ^ + 2 j ^ + 3 k ^ OB → = 3 i ^ - 4 j ^ + 5 k ^ OC → = - 2 i ^ + 3 j ^ - 7 k ^ Now, AB → = OB → - OA → AB → = 2 i ^ - 6 j ^ + 2 k ^ ⇒ | AB → | = | 2 i ^ - 6 j ^ + 2 k ^ | ⇒ AB → = 4 + 36 + 4 = 44 Consider, BC → = OC → - OB → = - 5 i ^ + 7 j ^ - 12 k ^ ⇒ | BC → | = | - 5 i ^ + 7 j ^ - 12 k ^ | = ( - 5 ) 2 + 7 2 + ( - 12 ) 2 = 218 Consider, AC → = OC → - OA → AC → = - 3 i ^ + j ^ - 10 k ^ ⇒ | AC → | = | - 3 i ^ + j ^ - 10 k ^ | = ( - 3 ) 2 + 1 2 + ( - 10 ) 2 = 9 + 1 + 100 = 110 ∴ The required perimeter of the triangle ABC is AB + BC + AC = 44 + 218 + 110 units.

Find the unit vector parallel to 3 a → → - 2 b → + 4 c → if a → = 3 i ^ - j ^ - 4 k ^ , b → = - 2 i ^ + 4 j ^ - 3 k ^ and c → = i ^ + 2 j ^ - k ^ .

Given that, the vectors are a → = 3 i ^ - j ^ - 4 k ^ b → = - 2 i ^ + 4 j ^ - 3 k ^ c → = i ^ + 2 j ^ - k ^ Consider, 3 a → - 2 b → + 4 c → = 3 ( 3 i ^ - j ^ - 4 k ^ ) - 2 ( - 2 i ^ + 4 j ^ - 3 k ^ ) + 4 ( i ^ + 2 j ^ - k ^ ) = 9 i ^ - 3 j ^ - 12 k ^ + 4 i ^ - 8 j ^ + 6 k ^ + 4 i ^ + 8 j ^ - 4 k ^ ∴ 3 a → - 2 b → + 4 c → = 17 i ^ - 3 j ^ - 10 k ^ The unit vector parallel to A B → is A B → A B → . The unit vector parallel to 3 a → - 2 b → + 4 c → is = 3 a → - 2 b → + 4 c → | 3 a → - 2 b → + 4 c → | = 17 i ^ - 3 j ^ - 10 k ^ | 17 i ^ - 3 j ^ - 10 k ^ | = 17 i ^ - 3 j ^ - 10 k ^ 17 2 + ( - 3 ) 2 + ( 10 ) 2 = 17 i ^ - 3 j ^ - 10 k ^ 398 Thus, the unit vector parallel to 3 a → - 2 b → + 4 c → is 17 i ^ - 3 j ^ - 10 k ^ 398 .

The position vector a → , b → , c → three points satisfy the relation 2 a → - 7 b → + 5 c → = 0 → . Are these points collinear?

Let A , B , C be the three points whose position vectors are a → , b → and c → OA → = a → , OB → = b → and OC → = c → . The position vectors satisfy the condition ⇒ 2 a → - 7 b → + 5 c → = 0 ⇒ 2 a → + 5 c → = 7 b → ⇒ 2 a → + 5 c → - 7 a → = 7 b → - 7 a → ⇒ 5 ( c → - a → ) = 7 ( b → - a → ) ⇒ 5 ( c → - a → ) = 7 ( b → - a → ) ⇒ c → - a → = 7 5 ( b → - a → ) c → - a → = λ ( b → - a → ) ......(i) AB → = OB → - OA → AB → = b → - a → AC → = OC → - OA → AC → = c → - a → ⇒ A C → = λ A B → [from(i)] A C → and A B → are parallel vectors and A is common point. ∴ Yes, A , B , C are collinear.

The position vectors of the point P , Q , R , S are i ^ + j ^ + k ^ , 2 i ^ + 5 j ^ , 3 i ^ + 2 j ^ - 3 k ^ and i ^ - 6 j ^ - k ^ respectively. Prove that the line PQ and RS are parallel.

Given that, the position vectors of the given points P , Q , R , S are OP → = i ^ + j ^ + k ^ OQ → = 2 i ^ + 5 j ^ OQ → = 2 i ^ + 5 j ^ OR → = 3 i ^ + 2 j ^ - 3 k ^ OS → = i ^ - 6 j ^ - k ^ Now, PQ → = OQ → - OP → = ( 2 i ^ + 5 j ^ ) - ( i ^ + j ^ + k ^ ) = 2 i ^ + 5 j ^ - i ^ - j ^ - k ^ So, PQ → = i ^ + 4 j ^ - k ^ Consider, RS → = OS → - OR → = ( i ^ - 6 j ^ - k ^ ) - ( 3 i ^ + 2 j ^ - 3 k ^ ) = - 2 i ^ - 8 j ^ + 2 k ^ = - 2 ( i ^ + 4 j ^ - k ^ ) ∴ RS → = - 2 PQ → We know that, two vectors a → and b → are parallel vectors if a → = λ b → where λ is a scalar. ∴ RS → and PQ → are parallel vectors.

Find the value or values of m for which m ( i ^ + j ^ + k ^ ) is a unit vector.

Given that, the vector is m ( ı ^ + ȷ ^ + k ^ ) . Also, given that it is an unit vector. We know that, the magnitude of unit vector is 1 . ∴ | m ( i + j + k ) | = 1 m ± 1 2 + 1 2 + 1 2 = 1 m = ± 1 3 Hence, the values of m is ± 1 3 .

E M B I B E

Mathematics Standard 11 Vol II>Chapter 2 - Vector Algebra>EXERCISE

EXERCISE

Author:Tamil Nadu Board

Tamil Nadu Board Mathematics Solutions for EXERCISE

Simple step-by-step solutions to EXERCISE questions of Vector Algebra from Mathematics Standard 11 Vol II. Also get 3D topic explainers, cheat sheets, and unlimited doubts solving on EMBIBE.

Questions from EXERCISE with Hints & Solutions

MEDIUM

11th Tamil Nadu Board

IMPORTANT

Mathematics Standard 11 Vol II>Chapter 2 - Vector Algebra>EXERCISE>Q 11

If $\vec{a} = 2 \hat{i} + 3 \hat{j} - 4 \hat{k}, \vec{b} = 3 \hat{i} - 4 \hat{j} - 5 \hat{k}$ and $\vec{c} = - 3 \hat{i} + 2 \hat{j} + 3 \hat{k}$ , find the magnitude and direction cosines of $\vec{a} + \vec{b} + \vec{c}$

MEDIUM

11th Tamil Nadu Board

IMPORTANT

Mathematics Standard 11 Vol II>Chapter 2 - Vector Algebra>EXERCISE>Q 11

If $\vec{a} = 2 \hat{i} + 3 \hat{j} - 4 \hat{k}, \vec{b} = 3 \hat{i} - 4 \hat{j} - 5 \hat{k}$ and $\vec{c} = - 3 \hat{i} + 2 \hat{j} + 3 \hat{k}$ , find the magnitude and direction cosines of $3 \vec{a} - 2 \vec{b} + 5 \vec{c}$

EASY

11th Tamil Nadu Board

IMPORTANT

Mathematics Standard 11 Vol II>Chapter 2 - Vector Algebra>EXERCISE>Q 12

The position vectors of the vertices of a triangle are $\hat{i} + 2 \hat{j} + 3 \hat{k}, 3 \hat{i} - 4 \hat{j} + 5 \hat{k}$ and $- 2 \hat{i} + 3 \hat{j} - 7 \hat{k}$ . Find the perimeter of the triangle.

EASY

11th Tamil Nadu Board

IMPORTANT

Mathematics Standard 11 Vol II>Chapter 2 - Vector Algebra>EXERCISE>Q 13

Find the unit vector parallel to $\vec{3 \vec{a}} - 2 \vec{b} + 4 \vec{c}$ if $\vec{a} = 3 \hat{i} - \hat{j} - 4 \hat{k}, \vec{b} = - 2 \hat{i} + 4 \hat{j} - 3 \hat{k}$ and $\vec{c} = \hat{i} + 2 \hat{j} - \hat{k}$ .

EASY

11th Tamil Nadu Board

IMPORTANT

Mathematics Standard 11 Vol II>Chapter 2 - Vector Algebra>EXERCISE>Q 14

The position vector $\vec{a}, \vec{b}, \vec{c}$ three points satisfy the relation $2 \vec{a} - 7 \vec{b} + 5 \vec{c} = \vec{0}$ . Are these points collinear?

EASY

11th Tamil Nadu Board

IMPORTANT

Mathematics Standard 11 Vol II>Chapter 2 - Vector Algebra>EXERCISE>Q 15

The position vectors of the point $P, Q, R, S$ are $\hat{i} + \hat{j} + \hat{k}, 2 \hat{i} + 5 \hat{j}, 3 \hat{i} + 2 \hat{j} - 3 \hat{k}$ and $\hat{i} - 6 \hat{j} - \hat{k}$ respectively. Prove that the line $PQ$ and $RS$ are parallel.