MCQ
Let $\vec{a}=\hat{i}+\hat{j}+2 \hat{k}, \vec{b}=2 \hat{i}-3 \hat{j}+\hat{k}$ and $\overrightarrow{ c }=\hat{ i }-\hat{ j }+\hat{ k }$ be three given vectors. Let $\vec{v}$ be a vector in the plane of $\vec{a}$ and $\overrightarrow{ b }$ whose projection on $\overrightarrow{ c }$ is $\frac{2}{\sqrt{3}}$. If $\overrightarrow{ v } . \hat{ j }=7$, then $\overrightarrow{ v } \cdot(\hat{ i }+\hat{ k })$ is equal to
  • A
    $6$
  • B
    $7$
  • C
    $8$
  • $9$

Answer

Correct option: D.
$9$
d
$\overrightarrow{ v }=\lambda \overrightarrow{ a }+\mu \overrightarrow{ b }$

$\vec{v}=\lambda(1,1,2)+\mu(2,-3,1)$

$\vec{v}=(\lambda+2 \mu, \lambda-3 \mu, 2 \lambda+\mu)$

$\overrightarrow{ v } \cdot \hat{ j }=7$

$\lambda-3 \mu=7$

$\overrightarrow{ v } \cdot \frac{\overrightarrow{ c }}{|\overrightarrow{ c }|}=\frac{2}{\sqrt{3}}$

$\overrightarrow{ V } \cdot \overrightarrow{ C }=2$

$\lambda+2 \mu-\lambda+3 \mu+2 \lambda+\mu=2$

$2 \lambda+6 \mu=2$

$\lambda+3 \mu=1$

$\lambda-3 \mu=7$

$2 \lambda=8$

$\lambda=4$

$\mu=-1$

We get $\quad \overrightarrow{ v }=(2,7,7)$

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