Scalar Triple Product and Planes
DAILY_CHALLENGE
Grade None
Question:
Let $\overrightarrow{OP}=\dfrac{\alpha-1}{\alpha}\hat{i}+\hat{j}+\hat{k}$, $\overrightarrow{OQ}=\hat{i}+\dfrac{\beta-1}{\beta}\hat{j}+\hat{k}$ and $\overrightarrow{OR}=\hat{i}+\hat{j}+\dfrac{1}{2}\hat{k}$ be three vectors, where $\alpha,\beta\in\mathbb{R}-\{0\}$ and $O$ denotes the origin. If $\left(\overrightarrow{OP}\times\overrightarrow{OQ}\right)\cdot\overrightarrow{OR}=0$ and the point $(\alpha,\beta,2)$ lies on the plane $3x+3y-z+l=0$, then the value of $l$ is ___.
Step-by-Step Solution
Key Concept: Scalar triple product zero implies coplanarity; then substitute into plane equation
$\overrightarrow{OP}\times\overrightarrow{OQ}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\\frac{\alpha-1}{\alpha}&1&1\\1&\frac{\beta-1}{\beta}&1\end{vmatrix}=\left(\frac{1}{\beta},\,\frac{1}{\alpha},\,\frac{(\alpha-1)(\beta-1)}{\alpha\beta}-1\right)$
Dot with $\overrightarrow{OR}=(1,1,\frac{1}{2})$:
$$\frac{1}{\beta}+\frac{1}{\alpha}+\frac{1}{2}\left(\frac{(\alpha-1)(\beta-1)}{\alpha\beta}-1\right)=0$$
Multiply by $2\alpha\beta$:
$$2\alpha+2\beta+(\alpha-1)(\beta-1)-\alpha\beta=0$$
$$2\alpha+2\beta+\alpha\beta-\alpha-\beta-1-\alpha\beta=0$$
$$\alpha+\beta-1=0\Rightarrow\alpha+\beta=1.$$
$(\alpha,\beta,2)$ on $3x+3y-z+l=0$:
$$3\alpha+3\beta-2+l=0\Rightarrow3(1)-2+l=0\Rightarrow l=-1.$$
Correct Answer: -1