Name: Equation of a Line in 3D (Symmetric Form)
Uploaded: 2019-07-19
Duration: 11 min 8 s
Description: We have seen the equation of a line in 2D. Here, in this video, we will see the equation of a line in 3D and also see why vectors play an important role in 3...

Question

If

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

and

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

are two lines, then find the equation of acute angle bisector of two lines.

Accepted Answer

Given data $\vec{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (\hat{i} - \hat{j} - \hat{k}) . . . line 1$

and $\vec{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + μ (\hat{i} + \hat{j} - \hat{k}) . . . line 2$

Position vector of $P (1, 2, 3)$

and let $\vec{a} = \hat{i} - \hat{j} + \hat{k}, \vec{b} = (\hat{i} + \hat{j} - \hat{k})$

We know that:

$\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| cosθ$

So, $cosθ = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$

$cosθ = \frac{(\hat{i} - \hat{j} + \hat{k}) (\hat{i} + \hat{j} - \hat{k})}{[\sqrt{1 + 1 + 1}] [\sqrt{1 + 1 + 1}]}$

$cosθ = \frac{1 - 1 - 1}{\sqrt{9}} = \frac{- 1}{3}$

Since $(1)$ is acute angle because $cosθ$ is $- ve$

So direction of acute angle bisector.

$= \frac{(\hat{i} - \hat{j} + \hat{k})}{\sqrt{3}} \cdot \frac{(\hat{i} + \hat{j} - \hat{k})}{\sqrt{3}} = \frac{- 2 \hat{j} + 2 \hat{k}}{3}$

Acute angle bisector $= \hat{i} + 2 \hat{j} + 3 \hat{k} + δ (\frac{- 2 \hat{j} + 2 \hat{k}}{\sqrt{3}})$

Let $δ$ is any constant

So $\vec{r} = \hat{i} + 2 \hat{j} + 3 \hat{k} + t (- \hat{j} + \hat{k}) (let t = \frac{- 2 δ}{\sqrt{3}})$

So acute angle bisector $\vec{r} = \hat{i} + 2 \hat{j} + 3 \hat{k} + t (- \hat{j} + \hat{k})$

If $\vec{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (\hat{i} - \hat{j} + \hat{k})$ and $\vec{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + μ (\hat{i} + \hat{j} - \hat{k})$ are two lines, then find the equation of acute angle bisector of two lines.

Important Questions on Three Dimensional Geometry

Learn and Prepare for any exam you want !

Ifr→=i^+2j^+3k^+λi^-j^+k^ and r→=i^+2j^+3k^+μi^+j^-k^ are two lines, then find the equation of acute angle bisector of two lines.

Important Questions on Three Dimensional Geometry

Learn and Prepare for any exam you want !

If $\vec{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + λ (\hat{i} - \hat{j} + \hat{k})$ and $\vec{r} = (\hat{i} + 2 \hat{j} + 3 \hat{k}) + μ (\hat{i} + \hat{j} - \hat{k})$ are two lines, then find the equation of acute angle bisector of two lines.