Question

If the vectors $\vec{a} = \hat{i}-2\hat{j}+\hat{k}$, $\vec{b}=2\hat{i}-5\hat{j} + p\hat{k}$ and $\vec{c} = 5\hat{i}-9\hat{j}+ 4\hat{k}$ are coplanar, then the value of p is

• A. -1/3
• B. 3
• C. -3
• D. 1/3

Answer 1

Answer: (B)

3

Answer 2

For three vectors to be coplanar, their scalar triple product must be zero. That is, $\vec{a} \cdot (\vec{b} \times \vec{c}) = 0$.

Given: $\vec{a} = (1, -2, 1)$ $\vec{b} = (2, -5, p)$ $\vec{c} = (5, -9, 4)$

First, compute $\vec{b} \times \vec{c}$:

$\vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -5 & p \\ 5 & -9 & 4 \end{vmatrix} = \hat{i}\Big((-5)(4) - (p)(-9)\Big) - \hat{j}\Big(2\cdot4 - p\cdot5\Big) + \hat{k}\Big(2(-9) - (-5)(5)\Big)$ $= \hat{i}(-20+9p) - \hat{j}(8-5p) + \hat{k}(-18+25)$ $= (9p-20)\hat{i} + (5p-8)\hat{j} + 7\hat{k}$.

Next, compute the scalar triple product:

$\vec{a} \cdot (\vec{b} \times \vec{c}) = 1\cdot(9p-20) + (-2)\cdot(5p-8) + 1\cdot7$ $= 9p - 20 - 10p + 16 + 7 = -p + 3$.

Setting the scalar triple product to zero:

$-p + 3 = 0 \implies p = 3$.

Therefore, the value of $p$ is 3.