In terms of unit vectors i, j and k, $a = 2, - 1, 1; b = 1, 3, - 2; c = - 2, 1, - 3$ and $d = 3, 2, 5.$ Determine the scalars x, y, z such that $d = x a + y b + z c .$

x = - 1, y = - 2, z = 3

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x = 1, y = 2, z = - 3

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x = 2, y = - 1, z = 3

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x = - 2, y = 1, z = - 3

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Solution

The correct option is D $x = - 2, y = 1, z = - 3$
Substitute the values and equate the coefficients $i, j, k$ on both sides.
Thus we have the following equations:
$2 x + y - 2 z = 3, - x + 3 y + z = 2, x - 2 y - 3 z = 5$
Sloving these, we get $x = - 2, y = 1, z = - 3$

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