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MHTCET 13 May 2023 Shift 2 Solved Paper

Section: Mathematics

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Question : 62 of 150

Marks: +1, -0

a = 2\hat{i} + 3\hat{j} + 4\hat{k}, b = \hat{i} - 2\hat{j} + \hat{k}, c = \hat{i} + \hat{j} - \hat{k}

are three vectors. For a vector

r

r \times a = b

r \cdot c = 3, |r|

is

$\sqrt{55}$
$\sqrt{138}$
$\sqrt{155}$
$\sqrt{170}$

Solution:

We have,

a = 2\hat{i} + 3\hat{j} + 4\hat{k}

b = \hat{i} - 2\hat{j} + \hat{k} \text{ and } c = \hat{i} + \hat{j} - \hat{k}

Let

r = x\hat{i} + y\hat{j} + z\hat{k}

Since,

r \times a = b

Since,

r \times a = b

\Rightarrow \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ x & y & z \\ 2 & 3 & 4 \end{vmatrix} = \hat{i} - 2\hat{j} + \hat{k}

\Rightarrow \hat{i}(4y - 3z) - \hat{j}(4x - 2z) + \hat{k}(3x - 2y) = \hat{i} - 2\hat{j} + \hat{k}

\Rightarrow 4y-3z=1 ; 4x-2z=2 ; \text{ and } 3x-2y=1 \} \dots (i)

Also,

r \cdot c = 3

\Rightarrow (x\hat{i} + y\hat{j} + z\hat{k}) \cdot (\hat{i} + \hat{j} - \hat{k}) = 3

\Rightarrow x + y - z = 3 \ldots \text{ (ii)}

On solving Eqs. (i) and (ii), we get

x = 5, y = 7 \text{ and } z = 9

\therefore \ |r| = \sqrt{x^2 + y^2 + z^2}

= \sqrt{25 + 49 + 81}

= \sqrt{155}

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