Let a vector vec{a} be coplanar with vectors | vedclass

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(B) Since $\vec{a}$ is coplanar with $\vec{b}$ and $\vec{c}$,we can write $\vec{a} = \lambda \vec{b} + \mu \vec{c}$.Substituting the vectors: $\vec{a}

Vedclass · Accepted Answer

$-42$

Vedclass · Answer

(B) Since $\vec{a}$ is coplanar with $\vec{b}$ and $\vec{c}$,we can write $\vec{a} = \lambda \vec{b} + \mu \vec{c}$.Substituting the vectors: $\vec{a} = \lambda(2 \hat{i} + \hat{j} + \hat{k}) + \mu(\hat{i} - \hat{j} + \hat{k}) = (2\lambda + \mu) \hat{i} + (\lambda - \mu) \hat{j} + (\lambda + \mu) \hat{k}$.Given $\vec{a} \cdot \vec{d} = 0$,where $\vec{d} = 3 \hat{i} + 2 \hat{j} + 6 \hat{k}$:$3(2\lambda + \mu) + 2(\lambda - \mu) + 6(\lambda + \mu) = 0$$6\lambda + 3\mu + 2\lambda - 2\mu + 6\lambda + 6\mu = 0$$14\lambda + 7\mu = 0 \implies \mu = -2\lambda$.Substituting $\mu$ back into $\vec{a}$: $\vec{a} = (2\lambda - 2\lambda) \hat{i} + (\lambda - (-2\lambda)) \hat{j} + (\lambda - 2\lambda) \hat{k} = 0 \hat{i} + 3\lambda \hat{j} - \lambda \hat{k}$.Given $|\vec{a}| = \sqrt{10}$,so $\sqrt{0^2 + (3\lambda)^2 + (-\lambda)^2} = \sqrt{10} \implies \sqrt{10\lambda^2} = \sqrt{10} \implies |\lambda| = 1$.Since $\vec{a}, \vec{b}, \vec{c}$ are coplanar,$[\vec{a} \vec{b} \vec{c}] = 0$.The expression becomes $[\vec{a} \vec{b} \vec{d}] + [\vec{a} \vec{c} \vec{d}] = [\vec{a} \vec{b} + \vec{c} \vec{d}]$.Using $\vec{b} + \vec{c} = 3 \hat{i} + 0 \hat{j} + 2 \hat{k}$ and $\vec{a} = 3\lambda \hat{j} - \lambda \hat{k}$:$[\vec{a} \vec{b} + \vec{c} \vec{d}] = \begin{vmatrix} 0 & 3\lambda & -\lambda \ 3 & 0 & 2 \ 3 & 2 & 6 \end{vmatrix} = 0(0 - 4) - 3\lambda(18 - 6) - \lambda(6 - 0) = -3\lambda(12) - 6\lambda = -42\lambda$.For $\lambda = 1$,the value is $-42$.

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