Question

Let \(\vec a,\vec b,\vec c\) be three vectors satisfying \(\vec a\times\vec b=\vec a\times\vec c\), \(|\vec a|=|\vec c|=1\), \(|\vec b|=4\) and \(|\vec b\times\vec c|=\sqrt{15}\). If \(\vec b-2\vec c=\lambda \vec a\), then \(\lambda\) equals

• A. \(1\)
• B. \(-1\)
• C. \(2\)
• D. \(-4\)

Hint:

If \(\vec a\times\vec x=0\), then \(\vec x\parallel\vec a\).

Answer 1

The Correct Option is B

Step 1: From \(\vec a\times\vec b=\vec a\times\vec c\), \[ \vec a\times(\vec b-\vec c)=0 \Rightarrow \vec b-\vec c \parallel \vec a \]
Step 2: Given \(\vec b-2\vec c=\lambda\vec a\).
Step 3: Using magnitudes: \[ |\vec b\times\vec c|=|\vec b||\vec c|\sin\theta \Rightarrow \sin\theta=\frac{\sqrt{15}}{4} \]
Step 4: Hence \(\cos\theta=\frac{1}{4}\).
Step 5: \[ |\vec b-2\vec c|^2=16+4-2(4)(2)\cos\theta=20-16\cdot\frac14=16 \] \[ \Rightarrow |\lambda|=4,\ \lambda < 0 \]