The vector(s) which is/are coplanar with vectors i + j +2 k and i +2 j + k , are perpendicular to the vector i + j + k is/are

Correct answer: A. j - k Let a= i + j +2 k , b= i +2 j + k and c= i + j + k A vector coplanar to a and b , and perpendicular to c . Now, λ(a × b) × c λ\ (a · c) b-(b · c) a\ λ\ (1+1+4)( i +2 j + k ) -(1+2+1)( i + j +2 k )\ λ\ 6 i +12 j +6 k -6 i -6 j -12 k \ λ\ 6 j -6 k \ 6 λ( j - k ) For λ=(1)/(6) Option (a) is correct. For λ=-(1)/(6) Option (d) is correct.

JEE Advanced 2011Paper 1

Exam: JEE Advanced
Year: 2011
Paper: Paper 1
Subject: Mathematics
Chapter: Vector Algebra

JEE Advanced 2011 · Paper 1 · Mathematics · Vector Algebra

The vector(s) which is/are coplanar with vectors i + j +2 k and i...

The vector(s) which is/are coplanar with vectors

\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}

and

\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}

, are perpendicular to the vector

\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}

is/are

Options

A $\hat{\mathbf{j}}-\hat{\mathbf{k}}$
B $-\hat{\mathbf{i}}+\hat{\mathbf{j}}$
C $\hat{\mathbf{i}}-\hat{\mathbf{j}}$
D $-\hat{\mathbf{j}}+\hat{\mathbf{k}}$

Correct answer

A. $\hat{\mathbf{j}}-\hat{\mathbf{k}}$

Step-by-step solution

Let

\mathbf{a}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}

and

\mathbf{c}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}

\therefore

A vector coplanar to

\mathbf{a}

and

\mathbf{b}

, and perpendicular to

\mathbf{c}

. Now,

\quad \lambda(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}

\Rightarrow \quad \lambda\{(\mathbf{a} \cdot \mathbf{c}) \mathbf{b}-(\mathbf{b} \cdot \mathbf{c}) \mathbf{a}\}

\Rightarrow \quad \lambda\{(1+1+4)(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}})

-(1+2+1)(\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}})\}

\Rightarrow \lambda\{6 \hat{\mathbf{i}}+12 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}-6 \hat{\mathbf{i}}-6 \hat{\mathbf{j}}-12 \hat{\mathbf{k}}\}

\Rightarrow \lambda\{6 \hat{\mathbf{j}}-6 \hat{\mathbf{k}}\} \Rightarrow 6 \lambda(\hat{\mathbf{j}}-\hat{\mathbf{k}})

For

\lambda=\frac{1}{6} \Rightarrow

Option (a) is correct. For

\lambda=-\frac{1}{6} \Rightarrow

Option (d) is correct.

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