Q38 / 100 · BITSAT 2023 (Memory Based Paper 2) · Mathematics
If the straight lines $\frac{x-1}{k}=\frac{y-2}{2}=\frac{z-3}{3}$ and $\frac{x-2}{3}=\frac{y-3}{k}=\frac{z-1}{2}$ intersect at a point, then the integer $\mathrm{k}$ is equal to
- Correct answerA
-5 - B
5 - C
2 - D
-2
Solution
When the two lines intersect then shortes distance between them is zero i.e.
$\begin{aligned}<br />& \frac{\left(\vec{a}_2-\vec{a}_1\right) \cdot \vec{b}_1 \times \vec{b}_2}{\left|\vec{b}_1 \times \vec{b}_2\right|}=0 \\<br />& \Rightarrow\left(\vec{a}_2-\vec{a}_1\right) \cdot \vec{b}_1 \times \vec{b}_2=0<br />\end{aligned}$
where $\vec{a}_1=\hat{i}+2 \hat{j}+3 \hat{k}, \vec{b}_1=k \hat{i}+2 \hat{j}+3 \hat{k}$
$\begin{aligned}<br />& \vec{a}_2=2 \hat{i}+3 \hat{j}+\hat{k}, \hat{b}_2=3 \hat{i}+k \hat{j}+2 \hat{k} \\<br />& \Rightarrow\left|\begin{array}{ccc}<br />1 & 1 & -2 \\<br />k & 2 & 3 \\<br />3 & k & 2<br />\end{array}\right|=0 \\<br />& \Rightarrow \quad 1(4-3 k)-1(2 k-9)-2\left(k^2-6\right)=0 \\<br />& \Rightarrow \quad-2 k^2-5 k+25=0 \Rightarrow k=-5 \text { or } \frac{5}{2}<br />\end{aligned}$
$\begin{aligned}<br />& \frac{\left(\vec{a}_2-\vec{a}_1\right) \cdot \vec{b}_1 \times \vec{b}_2}{\left|\vec{b}_1 \times \vec{b}_2\right|}=0 \\<br />& \Rightarrow\left(\vec{a}_2-\vec{a}_1\right) \cdot \vec{b}_1 \times \vec{b}_2=0<br />\end{aligned}$
where $\vec{a}_1=\hat{i}+2 \hat{j}+3 \hat{k}, \vec{b}_1=k \hat{i}+2 \hat{j}+3 \hat{k}$
$\begin{aligned}<br />& \vec{a}_2=2 \hat{i}+3 \hat{j}+\hat{k}, \hat{b}_2=3 \hat{i}+k \hat{j}+2 \hat{k} \\<br />& \Rightarrow\left|\begin{array}{ccc}<br />1 & 1 & -2 \\<br />k & 2 & 3 \\<br />3 & k & 2<br />\end{array}\right|=0 \\<br />& \Rightarrow \quad 1(4-3 k)-1(2 k-9)-2\left(k^2-6\right)=0 \\<br />& \Rightarrow \quad-2 k^2-5 k+25=0 \Rightarrow k=-5 \text { or } \frac{5}{2}<br />\end{aligned}$
Hence, the correct answer is option: 'A'.