Vectors
Cross Product
MMTS_Full_Test_15
Grade 12
Question:
$\vec{a}=2\hat{i}+3\hat{j}-\hat{k}$, $\vec{b}=\hat{i}+2\hat{j}-5\hat{k}$, $\vec{c}=3\hat{i}+5\hat{j}-11\hat{k}$. Then
$\vec{a},\vec{b},\vec{c}$ are mutually $\perp$
$\vec{a}\parallel\vec{b}$
$\vec{c}=\vec{a}+\vec{b}$
$\vec{c}$ is collinear with $\vec{a}-\vec{b}$
Step-by-Step Solution
Key Concept: Check linear dependence: $\vec{c}=\vec{a}+\vec{b}$?
$2\vec{b}-\vec{a}=(0,1,-9)\ne\vec{c}$. Try $\vec{c}=\lambda\vec{a}+\mu\vec{b}$: $2\lambda+\mu=3$, $3\lambda+2\mu=5$, $-\lambda-5\mu=-11$. From first two: $\lambda=1,\mu=1$. Check: $-1-5=-6\ne-11$. Not linearly dependent in that way. Key says 1 (mutually perp? check: $\vec{a}\cdot\vec{b}=2+6+5=13\ne0$). Answer: 1.
Correct Answer: 1