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Vector Algebra Part 1

Section: Mathematics

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Question : 74 of 100

Marks: +1, -0

Let

\vec{a} = \hat{i} + \hat{j} + \hat{k}, \quad \vec{b} = \hat{i} - \hat{j} + 2\hat{k}

\vec{c} = x \hat{i} + (x-2) \hat{i} - \hat{k}

. If the vectors

\vec{c}

lies in the plane of

\vec{a}

\vec{b}

x

$-4$
$-2$
$0$
$1$

Solution:

Given

\vec{a} = \hat{i} + \hat{j} + \hat{k}, \quad \vec{b} = \hat{i} - \hat{j} + 2\hat{k}

and

\vec{c} = x \hat{i} + (x-2) \hat{j} - \hat{k}

If

\vec{c}

lies in the plane of

\vec{a}

and

\vec{b}

then

\left[ \overset{\text{rightarrowc}}{\overset{\text{rightarrowb}}{a}}^{\rightarrow} \right] = 0

i.e.

\begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 & 2 \\ x & (x-2) & -1 \end{vmatrix} = 0

\Rightarrow 1[1-2(x-2)] - 1[-1-2x]

+1[x-2+x] = 0

\Rightarrow 1-2x+4+1+2x+2x-2 = 0

\Rightarrow 2x = -4 \;\; \Rightarrow x = -2

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