Find the magnitude of vector a → = 2 i ^ - 7 j ^ - 3 k ^ .

Given that, the vector is a → = 2 i ^ - 7 j ^ - 3 k ^ We know that, for a vector a → = x i ^ + y j ^ + z k ^ , the magnitude is a ⇀ = x 2 + y 2 + z 2 ⇒ a ⇀ = 2 2 + 7 2 + 3 2 = 4 + 49 + 9 = 62 Thus the magnitude of the given vector is 62

Prove that: a → - b → b → - c → c → - a → = 0

Consider, a → - b → , b → - c → , c → - a → = a → - b → . b → - c → × c → - a → = a → - b → . b → × c → - b → × a → - c → × c → + c → × a → = a → - b → . b → × c → + a → × b → - 0 + c → × a → ∵ m → × m → = 0 , m → × n → = - n → × m → = a → . b → × c → + a → . a → × b → + a → . c → × a → - b → b → × c → - b → . a → × b → - b → . c → × a → = a → , b → , c → + 0 + 0 - 0 - 0 - b → , c → , a → = a → , b → , c → - a → , b → , c → = 0 .

The force of 6 units works parallel to the vector 2 i ^ - 2 j ^ + k ^ and displaces one particle from i ^ + 2 j ^ + 3 k ^ to 5 i ^ + 3 j ^ + 7 k ^ . Find out the work-done by force.

Let r ¯ F = 2 i ^ - 2 j ^ + k ^ , r ¯ 1 = i ^ + 2 j ^ + 3 k ^ , r ¯ 2 = 5 i ^ + 3 j ^ + 7 k ^ Now, F ¯ = | F | r ^ F = | F | r ¯ F r ¯ F = 6 2 i ^ - 2 j ^ + k ^ 2 2 + 2 2 + 1 = 4 i ^ - 4 j ^ + 2 k ^ Now, W = F ¯ . ( r ¯ 1 - r ¯ 2 ) = 4 i ^ - 4 j ^ + 2 k ^ . 4 i ^ + j ^ + 4 k ^ = 16 - 4 + 8 = 20 W Thus, work done by force is 20 Watts.

Find the area of the parallelogram whose diagonals are a ¯ = 3 i + j - 2 k and b ¯ = i - 3 j + 4 k .

Given that, a parallelogram whose diagonals are a ¯ = 3 i + j - 2 k and b ¯ = i - 3 j + 4 k . Area of parallelogram whose diagonals are given is Δ = 1 2 | a → × b → | But a → × b → = i ^ j ^ k ^ 3 1 - 2 1 - 3 4 = - 2 i ^ - 14 j ^ - 10 k ^ Hence, Δ = 1 2 | a → × b → | = 1 2 4 + 196 + 100 = 5 3 square units.

Find the value of i ^ · ( j ^ × k ^ ) + j ^ · ( i ^ × k ^ ) + k ^ · ( i ^ × j ^ ) .

We know that: i ^ × j ^ = k ^ , j ^ × k ^ = i ^ , k ^ × i ^ = j ^ ⇒ j ^ × i ^ = - k ^ , k ^ × j ^ = - i ^ , i ^ × k ^ = - j ^ Also, i ^ . i ^ = j ^ · j ^ = k ^ · k ^ = 1 ∴ i ^ . ( j ^ × k ^ ) + j ^ . ( i ^ × k ^ ) + k ^ . ( i ^ × j ^ ) = i ^ . i ^ + j ^ · ( - j ^ ) + k ^ · k ^ = 1 - 1 + 1 = 1 Thus, the value of i ^ . ( j ^ × k ^ ) + j ^ · ( i ^ × k ^ ) + k ^ · ( i ^ × j ^ ) = 1

Find the value of i ^ · ( j ^ × k ^ ) + j ^ · ( i ^ × k ^ ) + k ^ · ( i ^ × j ^ ) .

We know that: i ^ × j ^ = k ^ , j ^ × k ^ = i ^ , k ^ × i ^ = j ^ ⇒ j ^ × i ^ = - k ^ , k ^ × j ^ = - i ^ , i ^ × k ^ = - j ^ Also, i ^ . i ^ = j ^ · j ^ = k ^ · k ^ = 1 ∴ i ^ . ( j ^ × k ^ ) + j ^ . ( i ^ × k ^ ) + k ^ . ( i ^ × j ^ ) = i ^ . i ^ + j ^ · ( - j ^ ) + k ^ · k ^ = 1 - 1 + 1 = 1 Thus, the value of i ^ . ( j ^ × k ^ ) + j ^ · ( i ^ × k ^ ) + k ^ · ( i ^ × j ^ ) = 1

Find the area of the parallelogram whose diagonals are a ¯ = 3 i + j - 2 k and b ¯ = i - 3 j + 4 k .

Given that, a parallelogram whose diagonals are a ¯ = 3 i + j - 2 k and b ¯ = i - 3 j + 4 k . Area of parallelogram whose diagonals are given is Δ = 1 2 | a → × b → | But a → × b → = i ^ j ^ k ^ 3 1 - 2 1 - 3 4 = - 2 i ^ - 14 j ^ - 10 k ^ Hence, Δ = 1 2 | a → × b → | = 1 2 4 + 196 + 100 = 5 3 square units.

The force of 6 units works parallel to the vector 2 i ^ - 2 j ^ + k ^ and displaces one particle from i ^ + 2 j ^ + 3 k ^ to 5 i ^ + 3 j ^ + 7 k ^ . Find out the work-done by force.

Let r ¯ F = 2 i ^ - 2 j ^ + k ^ , r ¯ 1 = i ^ + 2 j ^ + 3 k ^ , r ¯ 2 = 5 i ^ + 3 j ^ + 7 k ^ Now, F ¯ = | F | r ^ F = | F | r ¯ F r ¯ F = 6 2 i ^ - 2 j ^ + k ^ 2 2 + 2 2 + 1 = 4 i ^ - 4 j ^ + 2 k ^ Now, W = F ¯ . ( r ¯ 1 - r ¯ 2 ) = 4 i ^ - 4 j ^ + 2 k ^ . 4 i ^ + j ^ + 4 k ^ = 16 - 4 + 8 = 20 W Thus, work done by force is 20 Watts.

Prove that: a → - b → b → - c → c → - a → = 0

Consider, a → - b → , b → - c → , c → - a → = a → - b → . b → - c → × c → - a → = a → - b → . b → × c → - b → × a → - c → × c → + c → × a → = a → - b → . b → × c → + a → × b → - 0 + c → × a → ∵ m → × m → = 0 , m → × n → = - n → × m → = a → . b → × c → + a → . a → × b → + a → . c → × a → - b → b → × c → - b → . a → × b → - b → . c → × a → = a → , b → , c → + 0 + 0 - 0 - 0 - b → , c → , a → = a → , b → , c → - a → , b → , c → = 0 .

Find the magnitude of vector a → = 2 i ^ - 7 j ^ - 3 k ^ .

Given that, the vector is a → = 2 i ^ - 7 j ^ - 3 k ^ We know that, for a vector a → = x i ^ + y j ^ + z k ^ , the magnitude is a ⇀ = x 2 + y 2 + z 2 ⇒ a ⇀ = 2 2 + 7 2 + 3 2 = 4 + 49 + 9 = 62 Thus the magnitude of the given vector is 62

If a → = i ^ - 2 j ^ + 4 k ^ and b → = i ^ - 3 k ^ , then find the value of | b → × 2 a → |

Given that, a → = i ^ - 2 j ^ + 4 k ^ and b → = i ^ - 3 k ^ . 2 a → = 2 i ^ - 4 j ^ + 8 k ^ Consider, b → × 2 a → = i ^ j ^ k ^ 1 0 - 3 2 - 4 8 = i ^ 0 + 12 - j ^ 8 + 6 + k ^ - 4 - 0 = 12 i ^ - 14 j ^ - 4 k ^ Now, b → × 2 a → = 12 i ^ - 14 j ^ - 4 k ^ = 12 2 + 14 2 + 4 2 = 144 + 196 + 16 = 18 . 86

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 10 - Vector Algebra>Chhattisgarh Board-2018>Q 3

EASY

12th Chhattisgarh Board

IMPORTANT

Earn 100

Find the area of the parallelogram whose two adjacent sides are represented by the vectors $\vec{a} = 2 \hat{i} - \hat{j} + \hat{k}$ and $\vec{b} = 3 \hat{i} - 4 \hat{j} - \hat{k}$ .

Important Questions on Vector Algebra

EASY

12th Chhattisgarh Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 10 - Vector Algebra>Chhattisgarh Board-2018>Q 4

Three forces acting on a particle are in equilibrium. If the angle between first two forces is $90 °$ and angle between second and third is $120 °$ , then find the ratio of the forces.