Let a vector a be coplanar with vectors b=2 i+j+k and c=i-j+k . I… - Vidyadip

MCQ

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 $\vec{d}=3 \vec{i}+2 \hat{j}+6 \hat{k}$, and $|\vec{a}|=\sqrt{10} .$ Then a possible value of $[\overrightarrow{\mathrm{a}} \overrightarrow{\mathrm{b}} \overrightarrow{\mathrm{c}}]+[\overrightarrow{\mathrm{a}} \overrightarrow{\mathrm{b}} \vec{d}]+[\overrightarrow{\mathrm{a}} \vec{c} \vec{d}]$ is equal to:

A
$-40$
✓
$-42$
C
$-29$
D
$-38$

✓

Answer

Correct option: B.

$-42$

b
$\vec{a}=\lambda \vec{b}+\mu \vec{c}=\hat{i}(2 \lambda+\mu)+\hat{j}(\lambda-\mu)+\hat{k}(\lambda+\mu)$

$\vec{a} \cdot \vec{d}=0=3(2 \lambda+\mu)+2(\lambda-\mu)+6(\lambda+\mu)$

$\Rightarrow 14 \lambda+7 \mu=0 \Rightarrow \mu=-2 \lambda$

$\Rightarrow \vec{a}=(0) \hat{i}-3 \lambda \hat{j}+(-\lambda) \hat{k}$

$\Rightarrow|\vec{a}|=\sqrt{10}|\lambda|=\sqrt{10} \Rightarrow|\lambda|=1$

$\Rightarrow \lambda=1 \text { or }-1$

${[\vec{a} \vec{b} \vec{c}]=0 \quad \text { (as vectors are coplanar) }}$

$[\vec{a} \vec{b} \vec{c}]+[\vec{a} \vec{b} \vec{d}]+[\vec{a} \vec{c} \vec{d}]=\left[\begin{array}{lll}\vec{a} & \vec{b}+\vec{c} & \vec{d}\end{array}\right]$

$=\left|\begin{array}{ccc}0 & -3 \lambda & \lambda \\ 3 & 0 & 2 \\ 3 & 2 & 6\end{array}\right|$

$=3 \lambda(12)+\lambda(6)=42 \lambda=-42$

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