Question

Let \(\overset{⃗}{a} = 2\hat{i}+\hat{k},\overset{⃗}{b} = \hat{i}+\hat{j}+\hat{k}\), and \(\overset{⃗}{c} = 4\hat{i}-3\hat{j}+7\hat{k}\). If \(\overset{⃗}{r}\) is a vector such that \(\overset{⃗}{r}\times \overset{⃗}{b} = \overset{⃗}{c}\times \overset{⃗}{b}\) and \(\overset{⃗}{r}\cdot \overset{⃗}{a} = 0\), Then \(\overset{⃗}{r}\cdot \overset{⃗}{c} =\)

• A. \(-14\)
• B. \(34\)
• C. \(-7\)
• D. \(20\)

Hint:

From r cross b = c cross b we get r minus c parallel to b.

Answer 1

The Correct Option is B

Step 1: Approach
Express $\vec r\cdot\vec c$ using $\vec r=\vec c+\lambda\vec b$ directly.

Step 2: Expand
$\vec r\cdot\vec c=|\vec c|^2+\lambda\,\vec b\cdot\vec c$. Here $|\vec c|^2=16+9+49=74$ and $\vec b\cdot\vec c=4-3+7=8$.

Step 3: Find lambda
From $\vec r\cdot\vec a=0$: $15+3\lambda=0$, so $\lambda=-5$.

Step 4: Result
$74-5\times8=34$. Option (B).

Final Answer:
The vector r equals c - 5b, and its dot product with c is 34, option (B). \[ \boxed{34} \]