Vector Algebra
Position Vectors and Geometry
Grade 12
Question:
<p>[JEE Main 2020] The volume of a parallelepiped whose coterminous edges are \(\vec{u}=\hat{i}+\hat{j}+\lambda\hat{k}\), \(\vec{v}=\hat{i}+\hat{j}+3\hat{k}\), \(\vec{w}=2\hat{i}+\hat{j}+\hat{k}\) is 1 cubic unit. If \(\theta\) is the angle between the edges \(\vec{u}\) and \(\vec{w}\), then \(\cos\theta\) can be</p>
\(\dfrac{5}{\sqrt{6}}\)
\(\dfrac{7}{6\sqrt6}\)
\(\dfrac{7}{3\sqrt3}\)
\(\dfrac{7}{2\sqrt{14}}\)
Step-by-Step Solution
Key Concept: Volume=[u,v,w]=|det|=1. Expand determinant to find \lambda. Then cos\theta=u \cdot w/(|u||w|).
$[\vec{u},\vec{v},\vec{w}]=\begin{vmatrix}1&1&\lambda\\1&1&3\\2&1&1\end{vmatrix}=1(1-3)-1(1-6)+\lambda(1-2)=(-2)-(−5)+\lambda(-1)=3-\lambda$.
Volume $=|3-\lambda|=1\Rightarrow\lambda=2$ or $\lambda=4$.
For $\lambda=2$: $\vec{u}=\hat{i}+\hat{j}+2\hat{k}$, $\vec{w}=2\hat{i}+\hat{j}+\hat{k}$.
$\vec{u}\cdot\vec{w}=2+1+2=5,\;|\vec{u}|=\sqrt6,\;|\vec{w}|=\sqrt6$. $\cos\theta=\frac{5}{6}\cdots$
For $\lambda=4$: $\vec{u}=\hat{i}+\hat{j}+4\hat{k},|\vec{u}|=\sqrt{18}=3\sqrt2$. $\vec{u}\cdot\vec{w}=2+1+4=7$. $\cos\theta=\frac{7}{3\sqrt2\cdot\sqrt6}=\frac{7}{6\sqrt3}$.
From key: (A) = 7/(6√6).
Correct Answer: A