Q74 / 160 · TS EAMCET 2020 (14 Sep Shift 1) · Mathematics

$L_1$ is a line passing through the points with position vectors $\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$ and $4 \hat{\mathbf{i}}-3 \hat{\mathbf{k}}$. $L_2$ is a line passing through the points with position vectors $\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$ and $2 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}-5 \hat{\mathbf{k}}$. Then the distance between $L_1$ and $L_2$ is

  • A
    0
  • B
    $\frac{3}{4}$
  • Correct answer
    C
    $\frac{4}{3}$
  • D
    $\frac{2}{3}$

Solution

Equation of line $L_1$ is $\mathbf{r}=(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}-\hat{\mathbf{k}})+\lambda(3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})$ Equation of line $L_2$ is $\mathbf{r}=(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}})+\mu(\hat{\mathbf{i}}-6 \hat{\mathbf{j}}-4 \hat{\mathbf{k}})$ Distance between $L_1$ and $L_2$ $\begin{array}{r} D=\frac{(4 \hat{\mathbf{j}}) \cdot[(3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \times(\hat{\mathbf{i}}-6 \hat{\mathbf{j}}-4 \hat{\mathbf{k}})]}{\mid(3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \times(\hat{\mathbf{i}}-\hat{\mathbf{j}})-4 \hat{\mathbf{k}}) \mid} \\ (3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \times(\hat{\mathbf{i}}-6 \hat{\mathbf{j}}-4 \hat{\mathbf{k}})=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 3 & 2 & -2 \\ 1 & -6 & -4 \end{array}\right| \\ =-(20 \hat{\mathbf{i}}-10 \hat{\mathbf{j}}+20 \hat{\mathbf{k}}) \\ D=\left|\frac{4 \hat{\mathbf{j}} \cdot(-20 \hat{\mathbf{i}}+10 \hat{\mathbf{j}}-20 \hat{\mathbf{k}})}{\sqrt{(20)^2+(10)^2+(20)^2}}\right|=\frac{4}{3} \end{array}$

Hence, the correct answer is option: 'C'.

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