CUET Mathematics 2024 - The distance between the lines r=i-2 j+3 k+(2 i+3 j+6 k) and r=3 i-2 j+1 k+(4 i+6 j+12 k) is : | PYQs + Solutions

CUET Mathematics 2024

Algebra

Vector Algebra

Medium

The distance between the lines $\vec{r}=\hat{i}-2 \hat{j}+3 \hat{k}+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k})$ and $\vec{r}=3 \hat{i}-2 \hat{j}+1 \hat{k}+\mu(4 \hat{i}+6 \hat{j}+12 \hat{k})$ is :

\frac{\sqrt{28}}{7}

\frac{\sqrt{199}}{7}

\frac{\sqrt{328}}{7}

\frac{\sqrt{421}}{7}

✅ Correct Option: 3

Distance Between Two Lines in 3D Space

Given Information:

\newline

- Line 1:

\vec{r} = \hat{i} - 2\hat{j} + 3\hat{k} + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k})

\newline

- Line 2:

\vec{r} = 3\hat{i} - 2\hat{j} + 1\hat{k} + \mu(4\hat{i} + 6\hat{j} + 12\hat{k})

Identify the direction vectors and points on each line.

Line 1:

\newline

- Direction vector:

\vec{a} = 2\hat{i} + 3\hat{j} + 6\hat{k}

\newline

- Point on line:

\vec{r}_1 = \hat{i} - 2\hat{j} + 3\hat{k}

Line 2:

\newline

- Direction vector:

\vec{b} = 4\hat{i} + 6\hat{j} + 12\hat{k}

\newline

- Point on line:

\vec{r}_2 = 3\hat{i} - 2\hat{j} + 1\hat{k}

Check if the direction vectors are parallel.

\vec{b} = 4\hat{i} + 6\hat{j} + 12\hat{k} = 2(2\hat{i} + 3\hat{j} + 6\hat{k}) = 2\vec{a}

Since

\vec{b} = 2\vec{a}

, the lines are parallel.

Calculate the distance between parallel lines using the formula:

d = \dfrac{|(\vec{r}_2 - \vec{r}_1) \times \vec{a}|}{|\vec{a}|}

Find

\vec{r}_2 - \vec{r}_1

\newline

\vec{r}_2 - \vec{r}_1 = (3\hat{i} - 2\hat{j} + 1\hat{k}) - (\hat{i} - 2\hat{j} + 3\hat{k})

\newline

\vec{r}_2 - \vec{r}_1 = 2\hat{i} - 2\hat{k}

Calculate

(\vec{r}_2 - \vec{r}_1) \times \vec{a}

\newline

(\vec{r}_2 - \vec{r}_1) \times \vec{a} = (2\hat{i} - 2\hat{k}) \times (2\hat{i} + 3\hat{j} + 6\hat{k})

Using the cross product formula:

\newline

\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & -2 \\ 2 & 3 & 6 \end{vmatrix}

= \hat{i}(0 \times 6 - (-2) \times 3) - \hat{j}(2 \times 6 - (-2) \times 2) + \hat{k}(2 \times 3 - 0 \times 2)

\newline

= \hat{i}(0 + 6) - \hat{j}(12 + 4) + \hat{k}(6 - 0)

\newline

= 6\hat{i} - 16\hat{j} + 6\hat{k}

Calculate

|(\vec{r}_2 - \vec{r}_1) \times \vec{a}|

\newline

|6\hat{i} - 16\hat{j} + 6\hat{k}| = \sqrt{6^2 + (-16)^2 + 6^2} = \sqrt{36 + 256 + 36} = \sqrt{328}

Calculate

|\vec{a}|

\newline

|\vec{a}| = |2\hat{i} + 3\hat{j} + 6\hat{k}| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7

Calculate the distance:

d = \dfrac{|(\vec{r}_2 - \vec{r}_1) \times \vec{a}|}{|\vec{a}|} = \dfrac{\sqrt{328}}{7}