Vector Algebra
Scalar Triple Product
Grade 12
Question:
<p>[JEE Main 2021] Let \(\vec{a}=2\hat{i}+\hat{j}-2\hat{k}\) and \(\vec{b}=\hat{i}+\hat{j}\). Let \(\vec{c}\) be a vector such that \(\vec{a}\cdot\vec{c}=|\vec{c}|\), \(|\vec{c}-\vec{a}|=2\sqrt2\) and the angle between \(\vec{a}\times\vec{b}\) and \(\vec{c}\) is \(\dfrac\pi6\). Then \(|(\vec{a}\times\vec{b})\times\vec{c}|\) equals</p>
<li>3/2</li>
<li>2</li>
<li>3</li>
<li>4</li>
Step-by-Step Solution
Key Concept: Compute a \times b. Find |c| from given conditions. Then |(a \times b) \times c|=|a \times b||c|sin(\pi/6).
$\vec{a}\times\vec{b}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&1&-2\\1&1&0\end{vmatrix}=(0+2)\hat{i}-(-0+2)\hat{j}+(2-1)\hat{k}=2\hat{i}-2\hat{j}+\hat{k}$.
$|\vec{a}\times\vec{b}|=\sqrt{4+4+1}=3$.
From $\vec{a}\cdot\vec{c}=|\vec{c}|$ and $|\vec{c}-\vec{a}|^2=8$:
$|\vec{c}|^2-2\vec{a}\cdot\vec{c}+|\vec{a}|^2=8\Rightarrow|\vec{c}|^2-2|\vec{c}|+9=8\Rightarrow|\vec{c}|^2-2|\vec{c}|+1=0\Rightarrow(|\vec{c}|-1)^2=0\Rightarrow|\vec{c}|=1$.
$|(\vec{a}\times\vec{b})\times\vec{c}|=|\vec{a}\times\vec{b}||\vec{c}|\sin\frac\pi6=3\cdot1\cdot\frac12=\frac32$. Answer: 3/2
Correct Answer: 3/2