If a→=λ(i^+j^−k^),b→=μ(i^−j^+k^), and c→ are unit vectors . c→perpendicular to the vector a→ and coplanar with a→ and b→, then a unit vector d→ perpendicular to both a→ and c→ is

Let c→=xi^+yj^+zk^. since a,b,c are coplanar , soxyz11−11−11=0⇒0⋅x−2y−2z=0⇒y+z=0.Also c→ is perpendicular to a→, so x+y−z=0. Therefore x−2=y1=z−1 and x2+y2+z2=1 ( c→ being a unit vector). We have x=−26,y=16,z=−16.Thus c→=16(−2i^+j^−k^) a→×c→=i^j^k^11-1-21-1=(0i+3j+3k)d→=1|a→×c→|a→×c→ d→=(0i+3j+3k)9+9=12(j+k).

If a→=λ(i^+j^−k^),b→=μ(i^−j^+k^), and c→ are unit vectors . c→perpendicular to the vector a→ and coplanar with a→ and b→, then a unit vector d→ perpendicular to both a→ and c→ is

Let c→=xi^+yj^+zk^. since a,b,c are coplanar , soxyz11−11−11=0⇒0⋅x−2y−2z=0⇒y+z=0.Also c→ is perpendicular to a→, so x+y−z=0. Therefore x−2=y1=z−1 and x2+y2+z2=1 ( c→ being a unit vector). We have x=−26,y=16,z=−16.Thus c→=16(−2i^+j^−k^) a→×c→=i^j^k^11-1-21-1=(0i+3j+3k)d→=1|a→×c→|a→×c→ d→=(0i+3j+3k)9+9=12(j+k).

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If $\vec{a} = λ (\hat{i} + \hat{j} - \hat{k}), \vec{b} = μ (\hat{i} - \hat{j} + \hat{k})$ , and $\vec{c}$ are unit vectors . $\vec{c}$ perpendicular to the vector $\vec{a}$ and coplanar with $\vec{a}$ and $\vec{b}$ , then a unit vector $\vec{d}$ perpendicular to both $\vec{a}$ and $\vec{c}$ is

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$\frac{1}{\sqrt{6}} (i - 2 j + k)$

$\frac{1}{\sqrt{2}} (j + k)$

$\frac{1}{\sqrt{6}} (2 i - j + k)$

$\frac{1}{\sqrt{2}} (j - k)$

answer is B.

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Detailed Solution

Let $\vec{c} = x \hat{i} + y \hat{j} + z \hat{k}$ . since a,b,c are coplanar , so
$|\begin{array}{ccr} x & y & z \\ 1 & 1 & - 1 \\ 1 & - 1 & 1 \end{array}| = 0 \Rightarrow 0 \cdot x - 2 y - 2 z = 0 \Rightarrow y + z = 0$ .
Also $\vec{c}$ is perpendicular to $\vec{a}$ , so $x + y - z = 0$ . Therefore $\frac{x}{- 2} = \frac{y}{1} = \frac{z}{- 1}$ and $x^{2} + y^{2} + z^{2} = 1$ ( $\vec{c}$ being a unit vector). We have $x = - \frac{2}{\sqrt{6}}, y = \frac{1}{\sqrt{6}}, z = - \frac{1}{\sqrt{6}}$ .
Thus $\vec{c} = \frac{1}{\sqrt{6}} (- 2 \hat{i} + \hat{j} - \hat{k}) \vec{a} \times \vec{c} = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & - 1 \\ - 2 & 1 & - 1 \end{matrix}| = (0 i + 3 j + 3 k)$
$\vec{d} = \frac{1}{| \vec{a} \times \vec{c} |} \vec{a} \times \vec{c} \vec{d} = \frac{(0 i + 3 j + 3 k)}{\sqrt{9 + 9}} = \frac{1}{\sqrt{2}} (j + k) .$