A vector d→ is equally inclined to three vector a→=i^−j^+k^,b→=2i^+j^ and c→=3j^−2k^ . Let x→,y→ and z→ be three vectors in the plane of a→,b→;b→,c→;c→,a→ respectively. Then

r → ⋅ d → = 0 , where r → = λ x → + μ v → + δ z →

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A vector $\vec{d}$ is equally inclined to three vector $\vec{a} = \hat{i} - \hat{j} + \hat{k}, \vec{b} = 2 \hat{i} + \hat{j} and \vec{c} = 3 \hat{j} - 2 \hat{k}$ . Let $\vec{x}, \vec{y} and \vec{z}$ be three vectors in the plane of $\vec{a}, \vec{b}; \vec{b}, \vec{c}; \vec{c}, \vec{a}$ respectively. Then

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$\vec{x} \cdot \vec{d} = - 1$

$\vec{y} \cdot \vec{d} = 1$

$\vec{z} \cdot \vec{d} = 0$

$\vec{r} \cdot \vec{d} = 0, where \vec{r} = λ \vec{x} + μ \vec{v} + δ \vec{z}$

answer is C.

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Detailed Solution

Since $[\vec{a} \vec{b} \vec{c}] = 0, \vec{a}, \vec{b} and \vec{c}$ are coplanar vectors.

Further, since $\vec{d}$ is equally inclined to $\vec{a}, \vec{b} and \vec{c}$

$\begin{array}{l} \vec{d} \cdot \vec{a} = \vec{d} \cdot \vec{b} = \vec{d} \cdot \vec{c} = 0 \\ \vec{d} \cdot \vec{x} = \vec{d} \cdot \vec{y} = \vec{d} \cdot \vec{z} = 0 \\ \vec{d} \cdot \vec{r} = 0 \end{array}$

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