Let a vector

Question:

Let a vector $\overrightarrow{\mathrm{a}}$ be coplanar with vectors $\overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}}$ and $\overrightarrow{\mathrm{c}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}+\hat{\mathrm{k}}$. If $\overrightarrow{\mathrm{a}}$ is perpendicular to $\overrightarrow{\mathrm{d}}=3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}$, and $|\overrightarrow{\mathrm{a}}|=\sqrt{10}$. Then a possible value of $\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]+\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{d}\end{array}\right]+\left[\begin{array}{lll}\vec{a} & \vec{c} & \vec{d}\end{array}\right]$ is equal to :

1. $-42$

2. $-40$

3. $-29$

4. $-38$

Correct Option: 1

Solution:

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

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

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

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

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

$\Rightarrow \lambda=1$ or $-1$

$[\vec{a} \vec{b} \vec{c}]=0$

$[\overrightarrow{\mathrm{a}} \overrightarrow{\mathrm{b}} \overrightarrow{\mathrm{c}}]+[\overrightarrow{\mathrm{a}} \overrightarrow{\mathrm{b}} \overrightarrow{\mathrm{d}}]+[\overrightarrow{\mathrm{a}} \overrightarrow{\mathrm{c}} \overrightarrow{\mathrm{d}}]=[\overrightarrow{\mathrm{a}} \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}} \overrightarrow{\mathrm{d}}]$

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

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