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STD 12 - 10. vector algebra — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 10. vector algebra4 Marks
MCQ
Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=\hat{i}-\hat{j}+\hat{k}$ and $\vec{c}=\hat{i}-\hat{j}-\hat{k}$ be three vectors. A vector $\vec{v}$ in the plane of $\vec{a}$ and $\vec{b}$, whose projection on $\vec{c}$ is $\frac{1}{\sqrt{3}}$, is given by
  • A
    $\hat{i}-3 \hat{j}+3 \hat{k}$
  • B
    $-3 \hat{i}-3 \hat{j}-\hat{k}$
  • ✓
    $3 \hat{i}-\hat{j}+3 \hat{k}$
  • D
    $\hat{i}+3 \hat{j}-3 \hat{k}$
✓

Answer

Correct option: C.
$3 \hat{i}-\hat{j}+3 \hat{k}$
c
$\text { Let } \vec{v}=\lambda \vec{a}+\mu \vec{b}$

$\Rightarrow \vec{v}=(\lambda+\mu) \hat{i}+(\lambda-\mu) \hat{j}+(\lambda+\mu) \hat{k}$

$\text { Projection of } \vec{v} \text { on } \vec{c}=\frac{\vec{v} \cdot \vec{c}}{|\vec{c}|}=\frac{1}{\sqrt{3}}$

$\Rightarrow \frac{(\lambda+\mu)-(\lambda-\mu)-(\lambda+\mu)}{\sqrt{3}}=\frac{1}{\sqrt{3}}$

$\Rightarrow \mu-\lambda=1$

$\text { or } \mu=\lambda+1$

$\Rightarrow \vec{v}=(2 \lambda+1) \hat{i}-\hat{j}+(2 \lambda+1) \hat{k}$

$\text { For } \lambda=1, \vec{v}=3 \hat{i}-\hat{j}+3 \hat{k}$

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