Question

Let a vector $\vec{a}$ be coplanar with vectors $\vec{b}=2 \hat{i}+\hat{j}+\hat{k}$ and $\vec{c}=\hat{i}-\hat{j}+\hat{k}$. If $\vec{a}$ is perpendicular to $\vec{d}=3 \hat{i}+2 \hat{j}+6 \hat{k}$, and $|\vec{a}|=\sqrt{10}$. Then a possible equal to:

• A. -42
• B. -40
• C. -29
• D. -38

Answer 1

Answer: (A)

$\vec{a}=\lambda \vec{b}+\mu \vec{c}=\hat{i}(2 \lambda+\mu)+\hat{j}(\lambda-\mu)+\hat{k}(\lambda+\mu)$
$\vec{a} \cdot \vec{d}=0=3(2 \lambda+\mu)+2(\lambda-\mu)+6(\lambda+\mu)$
$\Rightarrow 14 \lambda+7 \mu=0 $
$\Rightarrow \mu=-2 \lambda$
$\Rightarrow \vec{a}=(0) \hat{i}-3 \lambda \hat{j}+(-\lambda) \hat{k}$
$\Rightarrow|\vec{a}|=\sqrt{10}|\lambda|=\sqrt{10} \Rightarrow|\lambda|=1$
$\Rightarrow \lambda=1$ or $-1$
${[\vec{a}\,\,\, \vec{b}\,\,\, \vec{c}]=0 }$
${[\vec{a} \,\,\,\vec{b} \,\,\,\vec{c}]+\begin{bmatrix}\vec{a} & \vec{b} & \vec{d}\end{bmatrix}+\begin{bmatrix}\vec{a} &\vec{c} &\vec{d}\end{bmatrix}=[\vec{a} \,\,\,\vec{b}+\vec{c} \,\,\,\vec{d}] }$
$=\begin{vmatrix}0 & -3 \lambda & \lambda \\ 3 & 0 & 2 \\ 3 & 2 & 6\end{vmatrix}$
$=3 \lambda(12)+\lambda(6)=42 \lambda=-42$