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Model Paper 4 — MATHS STD 12 Science — Question

Rajasthan BoardEnglish MediumSTD 12 ScienceMATHSModel Paper 43 Marks
Question
If $\vec{a}=(\hat{i}-\hat{j}), \vec{b}=(3 \hat{j}-\hat{k})$ and $\vec{c}=(7 \hat{i}-\hat{k})$, find a vector $\vec{d}$ which is perpendicular to both $\vec{a}$ and $\vec{b}$ and for which $\vec{c} \cdot \vec{d}=1$
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Answer

$\text { Let } \vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}$
$\vec{d} \perp \vec{a}, \vec{d} \cdot \vec{a}=0$
$ \Rightarrow\left(a_1 \hat{\imath}+a_2 \hat{\jmath}+a_3 \hat{x}\right)(\hat{i}-\hat{j})=0$
$\Rightarrow a_1-a_2=0 \ldots \text { (i) }$
$\vec{d} \perp \vec{b}, \vec{a} \cdot \vec{b}=0$
$ \Rightarrow\left(a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}\right)(3 j-\hat{k})=0$
$\Rightarrow 3 a_2-a_2=0 \ldots \text { (ii) }$
$\vec{d} \vec{c}=1 $
$\Rightarrow\left(a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k}\right)(7 \hat{i}-\hat{k})=1$
$\Rightarrow 7 a_1-a_3=1 \ldots \text { (iii) }$
Solving equation $(i)$ and $(ii)$ we get $3 a _1- a _3=0$.
Again solving equation $(iii) \ (iv)$ we get $a_1=\frac{1}{4}$
From equation $(i), \ a_1-a_2=0$ or $a_1=a_2=\frac{1}{4}$
From equation $(ii), 3 a _2- a _2=0$
$ \Rightarrow 3 \cdot \frac{1}{4}=a_3 $
$\Rightarrow a_3=\frac{3}{4}$
Hence, $\vec{d}=\frac{1}{4} \hat{i}+\frac{1}{4} \hat{j}+\frac{3}{4} \hat{k}$

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