JEE Advanced 2021 Paper 1 · Q19 · Scalar & Vector Products

Question

Let $\vec{u}, \vec{v}$ and $\vec{w}$ be vectors in three-dimensional space, where $\vec{u}$ and $\vec{v}$ are unit vectors which are not perpendicular to each other and $\vec{u} \cdot \vec{w} = 1$, $\vec{v} \cdot \vec{w} = 1$, $\vec{w} \cdot \vec{w} = 4$. If the volume of the parallelopiped, whose adjacent sides are represented by the vectors $\vec{u}, \vec{v}$ and $\vec{w}$, is $\sqrt{2}$, then the value of $|3\vec{u} + 5\vec{v}|$ is _____ .

SolveKar AI · Accepted Answer

Let $\vec{u}\cdot\vec{v}=t$. The Gram determinant equals the square of the parallelepiped volume: $[\vec{u}\,\vec{v}\,\vec{w}]^2=\det\begin{pmatrix}1&t&1\t&1&1\1&1&4\end{pmatrix}=(1)(4-1)-t(4t-1)+1(t-1)=3-4t^2+t+t-1=2+2t-4t^2$. Set $=2$: $-4t^2+2t=0\Rightarrow 2t(1-2t)=0$. Since $\vec{u},\vec{v}$ are not perpendicular, $t
e 0$, so $t=1/2$. Then $|3\vec{u}+5\vec{v}|^2=9|\vec{u}|^2+30(\vec{u}\cdot\vec{v})+25|\vec{v}|^2=9+30\cdot\dfrac{1}{2}+25=9+15+25=49$. So $|3\vec{u}+5\vec{v}|=7$.

JEE Advanced 2021 Paper 1 · Q19 · Scalar & Vector Products

Solution