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Question

# If $\stackrel{\to }{a}$, $\stackrel{\to }{b}$, $\stackrel{\to }{c}$ are non-zero, non-coplanar vectors, prove that the following vectors are coplanar: (i) $5\stackrel{\to }{a}+6\stackrel{\to }{b}+7\stackrel{\to }{c,}7\stackrel{\to }{a}-8\stackrel{\to }{b}+9\stackrel{\to }{c}\mathrm{and}3\stackrel{\to }{a}+20\stackrel{\to }{b}+5\stackrel{\to }{c}$ (ii) $\stackrel{\to }{a}-2\stackrel{\to }{b}+3\stackrel{\to }{c},-3\stackrel{\to }{b}+5\stackrel{\to }{c}\mathrm{and}-2\stackrel{\to }{a}+3\stackrel{\to }{b}-4\stackrel{\to }{c}$

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Solution

## (i) The three vectors are coplanar if one of them is expressible as a linear combination of the other two . Let $5\stackrel{\to }{a}+6\stackrel{\to }{b}+7\stackrel{\to }{c}=x\left(7\stackrel{\to }{a}-8\stackrel{\to }{b}+9\stackrel{⇀}{c}\right)+y\left(3\stackrel{\to }{a}+20\stackrel{\to }{b}+5\stackrel{\to }{c}\right).\phantom{\rule{0ex}{0ex}}=\stackrel{\to }{a}\left(7x+3y\right)+\stackrel{\to }{b}\left(-8x+20y\right)+\stackrel{\to }{c}\left(9x+5y\right).$ $⇒7x+3y=5,-8x+20y=6\mathrm{and}9x+5y=7.$ Solving first two of these equations, we get $x=\frac{1}{2},y=\frac{1}{2}$. Clearly, these values of x and y satisfies the third equation. Hence, the given vectors are coplanar. (ii) The three vectors are coplanar if one of them is expressible as a linear combination of the other two. Let $\stackrel{\to }{a}-2\stackrel{\to }{b}+3\stackrel{\to }{c}=x\left(-3\stackrel{\to }{b}+5\stackrel{\to }{c}\right)+y\left(-2\stackrel{\to }{a}+3\stackrel{\to }{b}-4\stackrel{\to }{c}\right).\phantom{\rule{0ex}{0ex}}=\stackrel{\to }{a}\left(-2y\right)+\stackrel{\to }{b}\left(-3x+3y\right)+\stackrel{\to }{c}\left(5x-4y\right).$ $⇒-2y=1,-3x+3y=-2\mathrm{and}5x-4y=3$ Solving first two of these equations, we get $x=\frac{1}{6},y=-\frac{1}{2}$. These values of x and y does not satisfy the third equation. Hence, the given vectors are not coplanar.

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