Find the value of λ for which a → and b → are perpendicular if a → = 7 i ^ - λ j ^ - 7 k ^ and b → = 4 i ^ + 5 j ^ - k ^

a → = 7 i ^ - λ j ^ - 7 k ^ b → = 4 i ^ + 5 j ^ - k ^ since these two vectors are perpendicular,their dot product is zero. ⇒ a → · b → = | a → | | b → | cos θ = | a → | | b → | cos π 2 = 0 ⇒ a → · b → = ( 7 i ^ - λ j ^ - 7 k ^ ) · ( 4 i ^ + 5 j ^ - k ^ ) = 0 ⇒ a → · b → = ( 7 × 4 ) + ( - λ × 5 ) + ( - 7 × - 1 ) = 0 ⇒ a → · b → = 28 - 5 λ + 7 = 0 ⇒ λ = 7

If a → = ( i ^ - j ^ ) , b → = ( 3 j ^ - k ^ ) and c → = ( 7 i ^ - k ^ ) , find a vector d → which is perpendicular to both a → and b → and for which c → · d → = 4 .

Let d → be the vector which is perpendicular to a → and b → then we have, d → = ± k · a → × b → a → × b → , where k is a scalar We have, a → × b → = a 2 b 3 - b 2 a 3 i ^ + a 3 b 1 - b 3 a 1 j ^ + a 1 b 2 - b 1 a 2 k ^ . . . . ( i ) Here, We have a → = i ^ - j ^ , b → = 0 i ^ + 3 j ^ - k ^ and c → = 7 i ^ - k ^ ⇒ a 1 = 1 , a 2 = - 1 , a 3 = 0 and b 1 = 0 , b 2 = 3 , b 3 = - 1 Thus, substituting the values of a 1 , a 2 , a 3 and b 1 , b 2 and b 3 in equation ( i ) we get ⇒ a → × b → = - 1 × - 1 - 3 × 0 i ^ + 0 × 0 - - 1 × 1 j ^ + 1 × 3 - 0 × - 1 k ^ ⇒ a → × b → = 1 i ^ + 1 j ^ + 3 k ^ ⇒ | a → × b → | = ( 1 ) 2 + ( 1 ) 2 + ( 3 ) 2 ⇒ | a → × b → | = 11 ⇒ d → = ± k · i ^ + j ^ + 3 k ^ 11 Given that c → . d → = 4 ⇒ 7 i ^ - k ^ · k × i ^ + j ^ + 3 k ^ 11 = 4 ⇒ 7 k - 3 k 11 = 4 ⇒ k = 11 ⇒ d → = i ^ + j ^ + 3 k ^ 11 × 11 Therefore, d → = i ^ + j ^ + 3 k ^

Show that the points A , B , C , D with position vectors 4 i ^ + 8 j ^ + 12 k ^ , 2 i ^ + 4 j ^ + 6 k ^ , 3 i ^ + 5 j ^ + 4 k ^ and 5 i ^ + 8 j ^ + 5 k ^ respectively are coplanar.

Let A → = 4 i ^ + 8 j ^ + 12 k ^ , B → = 2 i ^ + 4 j ^ + 6 k ^ , C → = 3 i ^ + 5 j ^ + 4 k ^ & D → = 5 i ^ + 8 j ^ + 5 k ^ Now, A B → = Position vector of B - Position vector of A = 2 i ^ + 4 j ^ + 6 k ^ - 4 i ^ + 8 j ^ + 12 k ^ = - 2 i ^ - 4 j ^ - 6 k ^ Similarly, A C → = 3 i ^ + 5 j ^ + 4 k ^ - 4 i ^ + 8 j ^ + 12 k ^ = - i ^ - 3 j ^ - 8 k ^ and A D → = 5 i ^ + 8 j ^ + 5 k ^ - 4 i ^ + 8 j ^ + 12 k ^ = i ^ - 7 k ^ If A , B , C , D are coplanar, A B → , A C → , A D → lie in the same plane and are coplanar and A B → A C → A D → = 0 Now, A B → A C → A D → = - 2 - 4 - 6 - 1 - 3 - 8 1 0 - 7 = - 2 21 - 0 + 4 7 - - 8 - 6 0 - - 3 = - 42 + 60 - 18 = 0 Hence, A , B , C , D are coplanar.

Find a → × b → , if a → = i ^ - 7 j ^ + 7 k ^ and b → = 3 i ^ - 2 j ^ + 2 k ^ .

Given, a → = i ^ - 7 j ^ + 7 k ^ and b → = 3 i ^ - 2 j ^ + 2 k ^ To find, a → × b → : Now, a → × b → = i ^ j ^ k ^ 1 - 7 7 3 - 2 2 ⇒ a → × b → = i ^ - 14 + 14 - j ^ 2 - 21 + k ^ - 2 + 21 ⇒ a → × b → = 19 j ^ + 19 k ^ ⇒ a → × b → = 19 2 + 19 2 = 19 2 .

Find a → × b → , if a → = i ^ - 7 j ^ + 7 k ^ and b → = 3 i ^ - 2 j ^ + 2 k ^ .

Given, a → = i ^ - 7 j ^ + 7 k ^ and b → = 3 i ^ - 2 j ^ + 2 k ^ To find, a → × b → : Now, a → × b → = i ^ j ^ k ^ 1 - 7 7 3 - 2 2 ⇒ a → × b → = i ^ - 14 + 14 - j ^ 2 - 21 + k ^ - 2 + 21 ⇒ a → × b → = 19 j ^ + 19 k ^ ⇒ a → × b → = 19 2 + 19 2 = 19 2 .

Show that the points A , B , C , D with position vectors 4 i ^ + 8 j ^ + 12 k ^ , 2 i ^ + 4 j ^ + 6 k ^ , 3 i ^ + 5 j ^ + 4 k ^ and 5 i ^ + 8 j ^ + 5 k ^ respectively are coplanar.

Let A → = 4 i ^ + 8 j ^ + 12 k ^ , B → = 2 i ^ + 4 j ^ + 6 k ^ , C → = 3 i ^ + 5 j ^ + 4 k ^ & D → = 5 i ^ + 8 j ^ + 5 k ^ Now, A B → = Position vector of B - Position vector of A = 2 i ^ + 4 j ^ + 6 k ^ - 4 i ^ + 8 j ^ + 12 k ^ = - 2 i ^ - 4 j ^ - 6 k ^ Similarly, A C → = 3 i ^ + 5 j ^ + 4 k ^ - 4 i ^ + 8 j ^ + 12 k ^ = - i ^ - 3 j ^ - 8 k ^ and A D → = 5 i ^ + 8 j ^ + 5 k ^ - 4 i ^ + 8 j ^ + 12 k ^ = i ^ - 7 k ^ If A , B , C , D are coplanar, A B → , A C → , A D → lie in the same plane and are coplanar and A B → A C → A D → = 0 Now, A B → A C → A D → = - 2 - 4 - 6 - 1 - 3 - 8 1 0 - 7 = - 2 21 - 0 + 4 7 - - 8 - 6 0 - - 3 = - 42 + 60 - 18 = 0 Hence, A , B , C , D are coplanar.

If a → = ( i ^ - j ^ ) , b → = ( 3 j ^ - k ^ ) and c → = ( 7 i ^ - k ^ ) , find a vector d → which is perpendicular to both a → and b → and for which c → · d → = 4 .

Let d → be the vector which is perpendicular to a → and b → then we have, d → = ± k · a → × b → a → × b → , where k is a scalar We have, a → × b → = a 2 b 3 - b 2 a 3 i ^ + a 3 b 1 - b 3 a 1 j ^ + a 1 b 2 - b 1 a 2 k ^ . . . . ( i ) Here, We have a → = i ^ - j ^ , b → = 0 i ^ + 3 j ^ - k ^ and c → = 7 i ^ - k ^ ⇒ a 1 = 1 , a 2 = - 1 , a 3 = 0 and b 1 = 0 , b 2 = 3 , b 3 = - 1 Thus, substituting the values of a 1 , a 2 , a 3 and b 1 , b 2 and b 3 in equation ( i ) we get ⇒ a → × b → = - 1 × - 1 - 3 × 0 i ^ + 0 × 0 - - 1 × 1 j ^ + 1 × 3 - 0 × - 1 k ^ ⇒ a → × b → = 1 i ^ + 1 j ^ + 3 k ^ ⇒ | a → × b → | = ( 1 ) 2 + ( 1 ) 2 + ( 3 ) 2 ⇒ | a → × b → | = 11 ⇒ d → = ± k · i ^ + j ^ + 3 k ^ 11 Given that c → . d → = 4 ⇒ 7 i ^ - k ^ · k × i ^ + j ^ + 3 k ^ 11 = 4 ⇒ 7 k - 3 k 11 = 4 ⇒ k = 11 ⇒ d → = i ^ + j ^ + 3 k ^ 11 × 11 Therefore, d → = i ^ + j ^ + 3 k ^

Find the value of λ for which a → and b → are perpendicular if a → = 7 i ^ - λ j ^ - 7 k ^ and b → = 4 i ^ + 5 j ^ - k ^

a → = 7 i ^ - λ j ^ - 7 k ^ b → = 4 i ^ + 5 j ^ - k ^ since these two vectors are perpendicular,their dot product is zero. ⇒ a → · b → = | a → | | b → | cos θ = | a → | | b → | cos π 2 = 0 ⇒ a → · b → = ( 7 i ^ - λ j ^ - 7 k ^ ) · ( 4 i ^ + 5 j ^ - k ^ ) = 0 ⇒ a → · b → = ( 7 × 4 ) + ( - λ × 5 ) + ( - 7 × - 1 ) = 0 ⇒ a → · b → = 28 - 5 λ + 7 = 0 ⇒ λ = 7

E M B I B E

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

EASY

12th Nagaland Board

IMPORTANT

Earn 100

Using vectors, find the area of the triangle formed by $A (1, - 1, 2)$ , $B (2, 1, 1)$ and $C (3, - 1, 2)$ .

Important Questions on Vector Algebra

EASY

12th Nagaland Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 10 - Vector Algebra>Nagaland Board-2019>Q 5

If

p \hat{i} + 3 \hat{j}

is a vector of magnitude

5

, then the value of

p

is

MEDIUM

12th Nagaland Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 10 - Vector Algebra>Nagaland Board-2019>Q 6

Find the value of

λ

for which

\vec{a}

and

\vec{b}

are perpendicular if

\vec{a} = 7 \hat{i} - λ \hat{j} - 7 \hat{k}

and

\vec{b} = 4 \hat{i} + 5 \hat{j} - \hat{k}

HARD

12th Nagaland Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 10 - Vector Algebra>Nagaland Board-2019>Q 7

If $\vec{a} = (\hat{i} - \hat{j}), \vec{b} = (3 \hat{j} - \hat{k}) and \vec{c} = (7 \hat{i} - \hat{k})$ , find a vector $\vec{d}$ which is perpendicular to both $\vec{a}$ and $\vec{b}$ and for which $\vec{c} \cdot \vec{d} = 4$ .