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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:
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Verified Answer
The correct answers are:
$\hat{\mathbf{j}}-\hat{\mathbf{k}}$
,
$-\hat{\mathbf{j}}+\hat{\mathbf{k}}$
$\hat{\mathbf{j}}-\hat{\mathbf{k}}$
,
$-\hat{\mathbf{j}}+\hat{\mathbf{k}}$
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.
$\therefore$ 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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