If the points whose position vectors are $2 i + j + k, 6 i - j + 2 k$ and $14 i - 5 j + p k$ are collinear, then the value of $p$ is?

$2$
$4$
$6$
$8$

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Solution

The correct option is B
$4$

Step 1: Apply the condition of collinearity
Given the three vectors $2 i + j + k, 6 i - j + 2 k$ and $14 i - 5 j + p k$ are collinear.
Thus, the determinant of the three vectors is zero.
$|\begin{matrix} 2 & 1 & 1 \\ 6 & - 1 & 2 \\ 14 & - 5 & p \end{matrix}| = 0$
$\Rightarrow$ $2 (- P + 10) - 1 (6 P - 28) + (- 30 + 14) = 0$
Step 2:Solve the above equation for $p$
$\Rightarrow$ $- 2 P + 20 - 6 P + 28 - 16 = 0$
$\Rightarrow$ $- 8 P + 32 = 0$
$\Rightarrow$ $P = 4$
Hence, option B is the correct option.

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2 i + j + k, 6 i - j + 2 k

14 i - 5 j + p k

2 ¯ i + ¯ j + ¯ ¯ ¯ k, 6 ¯ i - ¯ j + 2 ¯ ¯ ¯ k

14 ¯ i - 5 ¯ j + p ¯ ¯ ¯ k

p

3 ¯ i - 2 ¯ j - ¯ ¯ ¯ k, 2 ¯ i + 3 ¯ j - 4 ¯ ¯ ¯ k, - ¯ i + ¯ j + 2 ¯ ¯ ¯ k

4 ¯ i + 5 ¯ j + λ ¯ ¯ ¯ k

λ = \frac{- 146}{m}

m

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