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BITSAT 2023Paper 2
Exam
BITSAT
Year
2023
Paper
Paper 2
Subject
Mathematics
Chapter
Three Dimensional Geometry

BITSAT 2023 · Paper 2 · Mathematics · Three Dimensional Geometry

If the straight lines (x-1)/(k)=(y-2)/(2)=(z-3)/(3) and...

If the straight lines x−1k=y−22=z−33\frac{x-1}{k}=\frac{y-2}{2}=\frac{z-3}{3} and x−23=y−3k=z−12\frac{x-2}{3}=\frac{y-3}{k}=\frac{z-1}{2} intersect at a point, then the integer k\mathrm{k} is equal to

Options

  1. A
    -5
  2. B
    5
  3. C
    2
  4. D
    -2

Correct answer

A.
-5

Step-by-step solution

When the two lines intersect then shortes distance between them is zero i.e.(a⃗2−a⃗1)⋅b⃗1×b⃗2∣b⃗1×b⃗2∣=0⇒(a⃗2−a⃗1)⋅b⃗1×b⃗2=0\begin{aligned} \\ & \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 \\ \\ & \Rightarrow\left(\vec{a}_2-\vec{a}_1\right) \cdot \vec{b}_1 \times \vec{b}_2=0 \\ \end{aligned}where a⃗1=i^+2j^+3k^,b⃗1=ki^+2j^+3k^\vec{a}_1=\hat{i}+2 \hat{j}+3 \hat{k}, \vec{b}_1=k \hat{i}+2 \hat{j}+3 \hat{k}a⃗2=2i^+3j^+k^,b^2=3i^+kj^+2k^⇒∣11−2k233k2∣=0⇒1(4−3k)−1(2k−9)−2(k2−6)=0⇒−2k2−5k+25=0⇒k=−5 or  52\begin{aligned} \\ & \vec{a}_2=2 \hat{i}+3 \hat{j}+\hat{k}, \hat{b}_2=3 \hat{i}+k \hat{j}+2 \hat{k} \\ \\ & \Rightarrow\left|\begin{array}{ccc} \\ 1 & 1 & -2 \\ \\ k & 2 & 3 \\ \\ 3 & k & 2 \\ \end{array}\right|=0 \\ \\ & \Rightarrow \quad 1(4-3 k)-1(2 k-9)-2\left(k^2-6\right)=0 \\ \\ & \Rightarrow \quad-2 k^2-5 k+25=0 \Rightarrow k=-5 \text{\text{ or } } \frac{5}{2} \\ \end{aligned}$
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