Vector Algebra
Scalar Triple Product with Cross Product — Magnitude
nta_pyq_2024_apr
Grade 12
Question:
Let $\vec{a}=6\hat{i}+\hat{j}-\hat{k}$ and $\vec{b}=\hat{i}+\hat{j}$. If $\vec{c}$ is a vector such that $|\vec{c}|\geq6$, $\vec{a}\cdot\vec{c}=6|\vec{c}|$, $|\vec{c}-\vec{a}|=2\sqrt{2}$ and the angle between $\vec{a}\times\vec{b}$ and $\vec{c}$ is $60^\circ$, then $|(\vec{a}\times\vec{b})\times\vec{c}|$ is equal to:
$\dfrac{9}{2}(6-\sqrt{6})$
$\dfrac{3}{2}\sqrt{6}$
$\dfrac{9}{2}(6+\sqrt{6})$
$\dfrac{3}{2}\sqrt{3}$
Step-by-Step Solution
Key Concept: $|\vec{c}-\vec{a}|^2=|\vec{c}|^2+38-12|\vec{c}|=8\Rightarrow|\vec{c}|^2-12|\vec{c}|+30=0\Rightarrow|\vec{c}|=6+\sqrt{6}$ (taking $|\vec{c}|\geq6$). $\vec{a}\times\vec{b}=\hat{i}-\hat{j}+5\hat{k}$, $|\vec{a}\times\vec{b}|=\sqrt{27}=3\sqrt{3}$.
$|\vec{c}|=6+\sqrt{6}$. $|(\vec{a}\times\vec{b})\times\vec{c}|=\dfrac{9}{2}(6+\sqrt{6})$.
Correct Answer: 3