Question

Let \(\bar{a}=2\hat{i}+\hat{k},\ \bar{b}=\hat{i}+\hat{j}+\hat{k},\) and \(\bar{c}=4\hat{i}-3\hat{j}+7\hat{k}\). If \(\bar{r}\) is a vector such that \(\bar{r}\times \bar{b}=\bar{c}\times \bar{b}\) and \(\bar{r}\cdot \bar{a}=0\), then the value of the scalar product \(\bar{r}\cdot \hat{i}\) is:

• A. \(-1\)
• B. \(2\)
• C. \(0\)
• D. \(-3\)

Hint:

From r x b = c x b deduce that r - c is parallel to b, write r = c + lambda b and use r . a = 0 to find lambda.

Answer 1

The Correct Option is A

Step 1: Use the cross product condition
\(\bar{r}\times \bar{b}=\bar{c}\times \bar{b}\Rightarrow (\bar{r}-\bar{c})\times \bar{b}=\bar{0}\), so \(\bar{r}-\bar{c}=\lambda \bar{b}\).

Step 2: Write r
\(\bar{r}=(4+\lambda )\hat{i}+(\lambda -3)\hat{j}+(7+\lambda )\hat{k}\).

Step 3: Use r . a = 0
\(2(4+\lambda )+(7+\lambda )=15+3\lambda =0\Rightarrow \lambda =-5\).

Step 4: Find r . i
\(\bar{r}=-\hat{i}-8\hat{j}+2\hat{k}\), so \(\bar{r}\cdot \hat{i}=-1\).

Final Answer:
The scalar product \(\bar{r}\cdot \hat{i}\) is \(-1\). \[ \boxed{\text{(A) }-1} \]