Let a vector ${\overrightarrow a }$ be coplanar with vectors $\overrightarrow b = 2\widehat i + \widehat j + \widehat k$ and $\overrightarrow c = \widehat i - \widehat j + \widehat k$. If ${\overrightarrow a}$ is perpendicular to $\overrightarrow d = 3\widehat i + 2\widehat j + 6\widehat k$, and $\left| {\overrightarrow a } \right| = \sqrt {10}$. Then a possible value of $$[\matrix{ {\overrightarrow a } & {\overrightarrow b } & {\overrightarrow c } \cr } ] + [\matrix{ {\overrightarrow a } & {\overrightarrow b } & {\overrightarrow d } \cr } ] + [\matrix{ {\overrightarrow a } & {\overrightarrow c } & {\overrightarrow d } \cr } ]$$ is equal to :
Solution
$$\overrightarrow a = \lambda \overrightarrow b + \mu \overrightarrow c = \widehat i(2\lambda + \mu ) + \widehat j(\lambda - \mu ) + \widehat k(\lambda + \mu )$$<br><br>$$\overrightarrow a \,.\,\overrightarrow d = 0 = 3(2\lambda + \mu ) + 2(\lambda - \mu ) + 6(\lambda + \mu )$$<br><br>$\Rightarrow 14\lambda + 7\mu = 0 \Rightarrow \mu = - 2\lambda$<br><br>$$ \Rightarrow \overrightarrow a = (0)\widehat i - 3\lambda \widehat j + ( - \lambda )\widehat k$$<br><br>$$ \Rightarrow \left| {\overrightarrow a } \right| = \sqrt {10} \left| \lambda \right| = \sqrt {10} \Rightarrow \left| \lambda \right| = 1$$<br><br>$\lambda = 1$ or $- 1$<br><br>$[\overrightarrow a \overrightarrow b \overrightarrow c ]$ = 0<br><br>$$[\overrightarrow a \overrightarrow b \overrightarrow c ] + [\overrightarrow a \overrightarrow b \overrightarrow d ] + [\overrightarrow a \overrightarrow c \overrightarrow d ] = [\overrightarrow a \overrightarrow b + \overrightarrow c \overrightarrow d ]$$<br><br>$$ = \left| {\matrix{
0 & { - 3\lambda } & \lambda \cr
3 & 0 & 2 \cr
3 & 2 & 6 \cr
} } \right|$$<br><br>$= 3\lambda (12) + \lambda (6) = 42\lambda = - 42$
About this question
Subject: Mathematics · Chapter: Vector Algebra · Topic: Types of Vectors
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