Vector Algebra
Cross product magnitude via angle condition and vector triple product
nta_pyq_2025_apr
Grade 12
Question:
Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}$, $\vec{b}=2\hat{i}+2\hat{j}+\hat{k}$ and $\vec{d}=\vec{a}\times\vec{b}$. If $\vec{c}$ is a vector such that $\vec{a}\cdot\vec{c}=|\vec{c}|$, $|\vec{c}-2\vec{a}|^2=8$ and the angle between $\vec{d}$ and $\vec{c}$ is $\dfrac{\pi}{4}$, then $|10-3\vec{b}\cdot\vec{c}|+|\vec{d}\times\vec{c}|^2$ is equal to ________.
Step-by-Step Solution
Key Concept: First find $|\vec{c}|$ from $|\vec{c}-2\vec{a}|^2=8$ (it gives $|\vec{c}|=2$), then use the cross-product magnitude condition $|\vec{d}\times\vec{c}|^2=|\vec{d}|^2|\vec{c}|^2\sin^2(\pi/4)=4$ combined with the vector triple product to get a quadratic in $\vec{b}\cdot\vec{c}$.
$\vec{d}=\vec{a}\times\vec{b}=-\hat{i}+\hat{j}$, $|\vec{d}|=\sqrt{2}$.
$|\vec{c}-2\vec{a}|^2=|\vec{c}|^2+4|\vec{a}|^2-4\vec{a}\cdot\vec{c}=|\vec{c}|^2+12-4|\vec{c}|=8 \Rightarrow (|\vec{c}|-2)^2=0 \Rightarrow |\vec{c}|=2$.
$|\vec{d}\times\vec{c}|^2=|\vec{d}|^2|\vec{c}|^2\sin^2(\pi/4)=2\cdot4\cdot\tfrac{1}{2}=4$.
Using $\vec{d}\times\vec{c}=(\vec{a}\times\vec{b})\times\vec{c}=(\vec{a}\cdot\vec{c})\vec{b}-(\vec{b}\cdot\vec{c})\vec{a}$ and letting $x=\vec{b}\cdot\vec{c}$:
$4=(\vec{a}\cdot\vec{c})^2|\vec{b}|^2+x^2|\vec{a}|^2-2(\vec{a}\cdot\vec{c})x(\vec{a}\cdot\vec{b})=4\cdot9+3x^2-2\cdot2\cdot x\cdot5$
$4=36+3x^2-20x \Rightarrow 3x^2-20x+32=0 \Rightarrow x=4$ or $x=\dfrac{8}{3}$.
Both values give: $|10-3x|+4=|10-12|+4=6$ (for $x=\tfrac{8}{3}$) or $|10-12|+4=6$ (for $x=4$).
$|10-3\vec{b}\cdot\vec{c}|+|\vec{d}\times\vec{c}|^2=6$.
Correct Answer: 6