Vector Algebra
Perpendicular Vectors — Finding $(14\cos\theta)^2$
nta_pyq_2024_apr
Grade 12
Question:
For $\lambda>0$, let $\theta$ be the angle between the vectors $\vec{a}=\hat{i}+\lambda\hat{j}-3\hat{k}$ and $\vec{b}=3\hat{i}-\hat{j}+2\hat{k}$. If the vectors $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ are mutually perpendicular, then the value of $(14\cos\theta)^2$ is equal to
Step-by-Step Solution
Key Concept: $(\vec{a}+\vec{b})\perp(\vec{a}-\vec{b})\Rightarrow|\vec{a}|^2=|\vec{b}|^2$. $1+\lambda^2+9=9+1+4\Rightarrow\lambda^2=4\Rightarrow\lambda=2$.
$\lambda=2$. $14\cos\theta=-5$. $(14\cos\theta)^2=25$.
Correct Answer: 3