If the points whose position vectors are 2i^+j^+k^,6i^−j^+2k^ and 14i^−5j^+pk^ are collinear, then the value of p is

Let a→=2i^+j^+k^,b=6i^−j+2k^ and c→=14i−5j^+pk^ are collinear, therefore a→b→c→=0⇒2116−1214−5p=0 ⇒2−p+10−16p−28+1−30+14=0⇒−2p+20−6p+28−16=0⇒−8p+32=0⇒p=4

If the points whose position vectors are 2i^+j^+k^,6i^−j^+2k^ and 14i^−5j^+pk^ are collinear, then the value of p is

Let a→=2i^+j^+k^,b=6i^−j+2k^ and c→=14i−5j^+pk^ are collinear, therefore a→b→c→=0⇒2116−1214−5p=0 ⇒2−p+10−16p−28+1−30+14=0⇒−2p+20−6p+28−16=0⇒−8p+32=0⇒p=4

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

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Detailed Solution

Let $\vec{a} = 2 \hat{i} + \hat{j} + \hat{k}, b = 6 \hat{i} - j + 2 \hat{k}$ and $\vec{c} = 14 i - 5 \hat{j} + p \hat{k}$ are collinear, therefore $[\begin{matrix} \vec{a} & \vec{b} & \vec{c} \end{matrix}] = 0$
$\Rightarrow |\begin{matrix} 2 & 1 & 1 \\ 6 & - 1 & 2 \\ 14 & - 5 & p \end{matrix}| = 0$
$\Rightarrow 2 [- p + 10] - 1 [6 p - 28] + 1 [- 30 + 14] = 0$