Find the unit vectors perpendicular to both a → and b → when a → = i ^ + 3 j ^ - 2 k ^ and b → = - i ^ + 3 k ^

We know that, 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 ^ + 3 j ^ - 2 k ^ and b → = - i ^ + 0 j ^ + 3 k ^ ⇒ a 1 = 1 , a 2 = 3 , a 3 = - 2 and b 1 = - 1 , b 2 = 0 , b 3 = 3 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 → = ( 9 - 0 ) i ^ + ( 2 - 3 ) j ^ + ( 0 - ( - 3 ) ) k ^ ⇒ a → × b → = ( 9 ) i ^ + ( - 1 ) j ^ + ( 3 ) k ^ ⇒ | a → × b → | = ( 9 ) 2 + ( - 1 ) 2 + ( 3 ) 2 ⇒ | a → × b → | = 91 Let r → be the unit vector which is perpendicular to a → & b → then we have, r → = ± ( a → × b → ) a → × b → Therefore, r → = ± 9 i ^ - j ^ + 3 k ^ 91

Find the unit vectors perpendicular to both a → and b → when a → = i ^ + 3 j ^ - 2 k ^ and b → = - i ^ + 3 k ^

We know that, 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 ^ + 3 j ^ - 2 k ^ and b → = - i ^ + 0 j ^ + 3 k ^ ⇒ a 1 = 1 , a 2 = 3 , a 3 = - 2 and b 1 = - 1 , b 2 = 0 , b 3 = 3 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 → = ( 9 - 0 ) i ^ + ( 2 - 3 ) j ^ + ( 0 - ( - 3 ) ) k ^ ⇒ a → × b → = ( 9 ) i ^ + ( - 1 ) j ^ + ( 3 ) k ^ ⇒ | a → × b → | = ( 9 ) 2 + ( - 1 ) 2 + ( 3 ) 2 ⇒ | a → × b → | = 91 Let r → be the unit vector which is perpendicular to a → & b → then we have, r → = ± ( a → × b → ) a → × b → Therefore, r → = ± 9 i ^ - j ^ + 3 k ^ 91

Find the unit vectors perpendicular to both a → and b → when a → = 4 i ^ + 2 j ^ - k ^ and b → = i ^ + 4 j ^ - k ^

We know that 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 → = 4 i ^ + 2 j ^ - k ^ and b → = i ^ + 4 j ^ - k ^ ⇒ a 1 = 4 , a 2 = 2 , a 3 = - 1 and b 1 = 1 , b 2 = 4 , 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 → = ( 2 × - 1 - ( - 1 ) × 4 ) i ^ + ( - 1 × 1 - ( - 1 ) × 4 ) j ^ + ( 4 × 4 - 1 × 2 ) k ^ ⇒ a → × b → = 2 i ^ + 3 j ^ + 14 k ^ ⇒ | a → × b → | = ( 2 ) 2 + ( 3 ) 2 + ( 14 ) 2 ⇒ | a → × b → | = 209 Let r → be the unit vector which is perpendicular to a → & b → then we have, r → = ± ( a → × b → ) a → × b → Therefore, r → = ± 2 i ^ + 3 j ^ + 14 k ^ 209

Find the unit vectors perpendicular to the plane of the vectors a → = 2 i ^ - 6 j ^ - 3 k ^ and b = 4 i ^ + 3 j ^ - k ^

We know that 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 → = 2 i ^ - 6 j ^ - 3 k ^ and b → = 4 i ^ + 3 j ^ - k ^ ⇒ a 1 = 2 , a 2 = - 6 , a 3 = - 3 and b 1 = 4 , 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 → = ( - 6 × ( - 1 ) - 3 × ( - 3 ) ) i ^ + ( - 3 × 4 - ( - 1 ) × 2 ) j ^ + ( 2 × 3 - 4 × ( - 6 ) ) k ^ ⇒ a → × b → = 15 i ^ - 10 j ^ + 30 k ^ ⇒ | a → × b → | = ( 15 ) 2 + ( - 10 ) 2 + ( 30 ) 2 ⇒ | a → × b → | = 1225 = 25 × 49 Let r → be the unit vector which is perpendicular to a → & b → then we have, r → = ± ( a → × b → ) a → × b → Therefore, r → = ± 3 i ^ - 2 j ^ + 6 k ^ 7

Find a vector of magnitude 5 units , perpendicular to each of the vectors ( a → + b → ) and ( a → - b → ) , where a → = ( i ^ + j ^ + k ^ ) and b → = ( i ^ + 2 j ^ + 3 k ^ )

Given, a → = ( i ^ + j ^ + k ^ ) and b → = ( i ^ + 2 j ^ + 3 k ^ ) a → + b → = 2 i + 3 j + 4 k = l → a → - b → = 0 i - i - 2 k = m → We know that 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 l → = 2 i ^ + 3 j ^ + 4 k ^ and m → = 0 i ^ - j ^ - 2 k ^ ⇒ a 1 = 2 , a 2 = 3 , a 3 = 4 and b 1 = 0 , b 2 = - 1 , b 3 = - 2 Thus, substituting the values of a 1 , a 2 , a 3 and b 1 , b 2 and b 3 , in equation ( i ) we get ⇒ l → × m → = ( 3 × - 2 - - 1 × 4 ) i ^ + ( 4 × 0 - ( - 2 ) × 2 ) j ^ + ( 2 × - 1 - 0 × 3 ) k ^ ⇒ l → × m → = - 2 i ^ + 4 j ^ - 2 k ^ ⇒ l → × m → = - 2 2 + 4 2 + - 2 2 ⇒ l → × m → = 24 Let r → be the vector which is perpendicular to l → & m → then we have, r → = ± k · ( l → × m → ) l → × m → where k is a scalar. r → = ± k - i ^ + 2 j ^ - k ^ 6 Here, as r → is of magnitude 5 i.e k = 5 , Thus, r → = ± 5 - i ^ + 2 j ^ - k ^ 6

Find an angle between two vectors a ⇀ and b → with magnitudes 1 and 2 respectively and | a → × b → | = 3 .

We are given that a → = 1 and b → = 2 and | a → × b → | = 3 , We know that, a → × b → = a → · b → sinθ ⇒ a → · b → sinθ = 3 ⇒ 1 × 2 × sin θ = 3 ⇒ 2 sin θ = 3 ⇒ θ = sin - 1 3 2 Therefore, θ = π 2

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 → = 1 .

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 → = 1 ⇒ 7 i ^ - k ^ · k × i ^ + j ^ + 3 k ^ 11 = 1 ⇒ 7 k - 3 k 11 = 1 ⇒ k = 11 4 ⇒ d → = i ^ + j ^ + 3 k ^ 11 × 11 4 Therefore, d → = 1 4 i ^ + j ^ + 3 k ^

If a → = ( 4 i ^ + 5 j ^ - k ^ ) , b → = ( i ^ - 4 j ^ + k ^ ) and c → = ( 3 i ^ + j ^ - k ^ ) , find a vector d → which is perpendicular to both a → and b → and for which c → · d → = 21 .

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 ^ Here, We have a → = 4 i ^ + 5 j ^ - k ^ , b → = i ^ - 4 j ^ + k ^ and c → = 3 i ^ + j ^ - k ^ ⇒ a 1 = 4 , a 2 = 5 , a 3 = - 1 and b 1 = 1 , b 2 = - 4 , 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 → = 5 × 1 - - 4 × - 1 i ^ + - 1 × 1 - 1 × 4 j ^ + 4 × - 4 - 1 × 5 k ^ ⇒ a → × b → = ( 1 ) i ^ + ( - 5 ) j ^ + ( - 21 ) k ^ Let d → be the vector which is perpendicular to a → & b → then we have, d → = ± k · a → × b → a → × b → , where k is a scalar ⇒ | a → × b → | = ( 1 ) 2 + ( - 5 ) 2 + ( - 21 ) 2 ⇒ | a → × b → | = 467 d → = ± k · i ^ - 5 j ^ - 21 k ^ 467 Given that c → · d → = 21 ⇒ 3 i ^ + j ^ - k ^ · ± k × i ^ - 5 j ^ - 21 k ^ 467 = 21 ⇒ 19 k 467 = 21 ⇒ k = 21 467 19 Therefore, d → = i ^ - 5 j ^ - 21 k ^ 19 467 × 21 467 d → = 21 19 i ^ - 5 j ^ - 21 k ^

Prove that | a → × b → | = ( a → · b → ) tan θ , where θ is the angle between a → and b → .

Given, | a → × b → | = ( a → · b → ) tan θ We know that a → . b → = | a → | | b → | cos θ And a → × b → = | | a → | | b → | sin θ | So, tan θ = a → × b → a → · b → a → × b → = a → · b → tan θ Hence, proved.

E M B I B E

SENIOR SECONDARY SCHOOL MATHEMATICS FOR CLASS 12>Chapter 24 - Cross or Vector, Product of Vectors>EXERCISE>Q 5

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12th CBSE

IMPORTANT

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Find the unit vectors perpendicular to both $\vec{a}$ and $\vec{b}$ when $\vec{a} = \hat{i} + 3 \hat{j} - 2 \hat{k}$ and $\vec{b} = - \hat{i} + 3 \hat{k}$

Important Questions on Cross or Vector, Product of Vectors

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12th CBSE

IMPORTANT

SENIOR SECONDARY SCHOOL MATHEMATICS FOR CLASS 12>Chapter 24 - Cross or Vector, Product of Vectors>EXERCISE>Q 5

Find the unit vectors perpendicular to both $\vec{a}$ and $\vec{b}$ when $\vec{a} = 4 \hat{i} + 2 \hat{j} - \hat{k} and \vec{b} = \hat{i} + 4 \hat{j} - \hat{k}$

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IMPORTANT

SENIOR SECONDARY SCHOOL MATHEMATICS FOR CLASS 12>Chapter 24 - Cross or Vector, Product of Vectors>EXERCISE>Q 6

Find the unit vectors perpendicular to the plane of the vectors $\vec{a} = 2 \hat{i} - 6 \hat{j} - 3 \hat{k} and b = 4 \hat{i} + 3 \hat{j} - \hat{k}$

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12th CBSE

IMPORTANT

SENIOR SECONDARY SCHOOL MATHEMATICS FOR CLASS 12>Chapter 24 - Cross or Vector, Product of Vectors>EXERCISE>Q 7

Find a vector of magnitude 6 which is perpendicular to both the vectors $\vec{a} = 4 \hat{i} - \hat{j} + 3 \hat{k} and \vec{b} = - 2 \hat{i} + \hat{j} - 2 \hat{k}$ .

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12th CBSE

IMPORTANT

SENIOR SECONDARY SCHOOL MATHEMATICS FOR CLASS 12>Chapter 24 - Cross or Vector, Product of Vectors>EXERCISE>Q 8

Find a vector of magnitude $5 units$ , perpendicular to each of the vectors $(\vec{a} + \vec{b}) and (\vec{a} - \vec{b})$ , where $\vec{a} = (\hat{i} + \hat{j} + \hat{k}) and \vec{b} = (\hat{i} + 2 \hat{j} + 3 \hat{k})$

EASY

12th CBSE

IMPORTANT

SENIOR SECONDARY SCHOOL MATHEMATICS FOR CLASS 12>Chapter 24 - Cross or Vector, Product of Vectors>EXERCISE>Q 9

Find an angle between two vectors $\overset{⇀}{a}$ and $\vec{b}$ with magnitudes $1$ and $2$ respectively and $| \vec{a} \times \vec{b} | = \sqrt{3}$ .

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12th CBSE

IMPORTANT

SENIOR SECONDARY SCHOOL MATHEMATICS FOR CLASS 12>Chapter 24 - Cross or Vector, Product of Vectors>EXERCISE>Q 10

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 $\vec{c} \cdot \vec{d} = 1$ .

HARD

12th CBSE

IMPORTANT

SENIOR SECONDARY SCHOOL MATHEMATICS FOR CLASS 12>Chapter 24 - Cross or Vector, Product of Vectors>EXERCISE>Q 11

If $\vec{a} = (4 \hat{i} + 5 \hat{j} - \hat{k}), \vec{b} = (\hat{i} - 4 \hat{j} + \hat{k}) and \vec{c} = (3 \hat{i} + \hat{j} - \hat{k})$ , find a vector $\vec{d}$ which is perpendicular to both $\vec{a}$ and $\vec{b}$ and for which $\vec{c} \cdot \vec{d} = 21$ .