Find the shortest distance between the lines.
$\vec{r} = (4 \hat{i} - \hat{j}) + λ (\hat{i} - \hat{j} + 2 \hat{k}) + μ (2 \hat{i} + 4 \hat{j} - 5 \hat{k)}$

Solution

$As \vec{r} = (4 \hat{i} - \hat{j}) + λ (\hat{i} + 2 \hat{j} - 3 \hat{k}) and \vec{r} = (\hat{i} - \hat{j} + 2 \hat{k}) + μ (2 \hat{i} + 4 \hat{j} - 5 \hat{k}) are two lines . \vec{a} = 4 \hat{i} - \hat{j} \vec{b} = \hat{i} + 2 \hat{j} - 3 \hat{k} \vec{c} = \hat{i} - \hat{j} + 2 \hat{k} \vec{d} = 2 \hat{i} + 4 \hat{j} - 5 \hat{k} shortes distance, d = |\begin{matrix} \frac{(\vec{b} x \vec{d}) x (\vec{c} - \vec{a})}{| \vec{b} x \vec{d} |} \end{matrix}| | \vec{b} x \vec{d} | = |\begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 2 & 4 & - 5 \end{matrix}| = \hat{i} (2) - \hat{j} (1) + \hat{k} (0) = 2 \hat{i} - \hat{j} | \vec{b} x \vec{d} | = \sqrt{4 + 1} = \sqrt{5} (\vec{c} - \vec{a}) . (\vec{b} x \vec{d}) = - 6 + 0 + 0 = - 6 ∴ d = |\frac{- 6}{\sqrt{5}}| = \frac{6}{\sqrt{5}}$

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Three Dimensional Geometry

Find the shortest distance between the lines.
$\vec{r} = (4 \hat{i} - \hat{j}) + λ (\hat{i} - \hat{j} + 2 \hat{k}) + μ (2 \hat{i} + 4 \hat{j} - 5 \hat{k)}$

Some More Questions From Three Dimensional Geometry Chapter

If α, β, γ are the angles which a line makes with the axes, prove that sin²α + sin² β + sin²γ = 2.

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Three Dimensional Geometry

Find the shortest distance between the lines.r→ = (4i^ -j^) + λ (i^ - j^ + 2k^) + μ (2i^ + 4j^-5k)^

Some More Questions From Three Dimensional Geometry Chapter

If α, β, γ are the angles which a line makes with the axes, prove that sin2α + sin2 β + sin2γ = 2.

Mock Test Series

NCERT Sample Papers

Entrance Exams Preparation

Find the shortest distance between the lines.
$\vec{r} = (4 \hat{i} - \hat{j}) + λ (\hat{i} - \hat{j} + 2 \hat{k}) + μ (2 \hat{i} + 4 \hat{j} - 5 \hat{k)}$

If α, β, γ are the angles which a line makes with the axes, prove that sin²α + sin² β + sin²γ = 2.