Question

The vector equation of the line passing through the point \((1,2,-4)\) and perpendicular to the two the lines \(\frac{x-8}{3}=\frac{y+19}{-16}=\frac{z-10}{7}\) and \(\frac{x-5}{3}=\frac{y-33}{8}=\frac{5-z}{5}\) is...

• A. \(\bar{r}=\hat{i}+2\hat{j}-4\hat{k}+\lambda (5\hat{i}+4\hat{j}+7\hat{k})\)
• B. \(\bar{r}=\hat{i}+2\hat{j}-4\hat{k}+\lambda (2\hat{i}+3\hat{j}+6\hat{k})\)
• C. \(\bar{r}=\hat{i}+2\hat{j}-4\hat{k}+\lambda (3\hat{i}+7\hat{j}+13\hat{k})\)
• D. \(\bar{r}=\hat{i}+2\hat{j}-4\hat{k}+\lambda (4\hat{i}+10\hat{j}+9\hat{k})\)

Hint:

The direction of the required line is the cross product of the direction vectors of the two given lines.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
The required line is perpendicular to both lines, so its direction is the cross product of their directions.

Step 2: Direction vectors
First line: \(\vec d_1=3\hat i-16\hat j+7\hat k\).
Second line: \(\frac{5-z}{5}=\frac{z-5}{-5}\), so \(\vec d_2=3\hat i+8\hat j-5\hat k\).

Step 3: Cross product
\(\vec d_1\times\vec d_2=\hat i(80-56)-\hat j(-15-21)+\hat k(24+48)=24\hat i+36\hat j+72\hat k=12(2\hat i+3\hat j+6\hat k)\).

Step 4: Write the line
Through \((1,2,-4)\) with direction \(2\hat i+3\hat j+6\hat k\): \(\bar r=\hat i+2\hat j-4\hat k+\lambda(2\hat i+3\hat j+6\hat k)\).

Final Answer:
The line has direction \(2\hat i+3\hat j+6\hat k\). \[ \boxed{\text{(B) }\bar{r}=\hat{i}+2\hat{j}-4\hat{k}+\lambda (2\hat{i}+3\hat{j}+6\hat{k})} \]