The position vectors of three consecutive vertices of a parallelogram are $i + j + k, i + 3 j + 5 k$ and $7 i + 9 j + 11 k$ . The position vector of the fourth vertex is

6 (i + j + k)

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7 (i + j + k)

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2 j - 4 k

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6 i + 8 j + 10 k

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Solution

The correct option is B $7 (i + j + k)$
Let $A (\to a) = i + j + k$
$B (\to b) = i + 3 j + 5 k C (\to c) = 7 i + 9 j + 11 k$
$D (\to d)$ is to be determined
$A B C D$ is a parallelogram so $\frac{1}{2} (\to b + \to d) = \frac{1}{2} (\to a + \to c)$
$\to d = \to a + \to c - \to b = 8 i + 10 j + 12 k - (i + 3 j + 5 k) = 7 i + 7 j + 7 k = 7 (i + j + k)$

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