Question:

The shortest distance between the lines $\vec{r}=\left(8+3\lambda\right)\hat{i}-\left(9+16\lambda\right)\hat{j}+\left(10+7\lambda\right)\hat{k}$ and $\vec{r}=15\hat{i}+29\hat{j}+5\hat{k}+\mu\left(3\hat{i}-8\hat{j}-5\hat{k}\right)$ is

Updated On: Jul 7, 2022
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The Correct Option is B

Solution and Explanation

Given, equation of lines are $\vec{r}=\left(8+3\lambda\right)\hat{i}-\left(9+16\lambda\right)\hat{j}+\left(10+7\lambda\right)\hat{k}$ i.e., $\vec{r}=8\hat{i}-9\hat{j}+10\hat{k}+\lambda\left(3\hat{i}-16\hat{j}+7\hat{k}\right)$ and $\vec{r}=15\hat{i}+29\hat{j}+5\hat{k}+\mu\left(3\hat{i}+8\hat{j}-5\hat{k}\right)$ Here, $\vec{a}_{1}=8\hat{i}-9\hat{j}+10\hat{k}$, $\vec{b}_{1}=3\hat{i}-16\hat{j}+7\hat{k}$ $\vec{a}_{2}=15\hat{i}+29\hat{j}+5\hat{k}$, $\vec{b}_{2}=3\hat{i}+8\hat{j}-5\hat{k}$ $\vec{a}_{2}-\vec{a}_{1}=\left(15\hat{i}+29\hat{j}+5\hat{k}\right)-\left(8\hat{i}-9\hat{j}+10\hat{k}\right)$ $=7\hat{i}+38\hat{j}-5\hat{k}$ $\vec{b}_{1}\times\vec{b}_{2}=\hat{i}\left(80-56\right)-\hat{j}\left(-15-21\right)+\hat{k}\left(24+48\right)$ $=\left(24\hat{i}+36\hat{j}+72\hat{k}\right)$ $\therefore$ Shortest distance, $d=\left|\frac{\left(\vec{a}_{2}-\vec{a}_{1}\right)\cdot\left(\vec{b}_{1} \times \vec{b}_{2}\right)}{\left|\vec{b}_{1} \times \vec{b}_{2}\right|}\right|$ $=\left|\frac{\left(7\hat{i}+38\hat{j}-5\hat{k}\right)\cdot\left(24\hat{i}+36\hat{j}+72\hat{k}\right)}{\sqrt{\left(24\right)^{2}+\left(36\right)^{2}+\left(72\right)^{2}}}\right|$ $=\left|\frac{168+1368-360}{\sqrt{576+1296+5184}}\right|=\left|\frac{1176}{\sqrt{7056}}\right|$ $=\frac{1176}{84}=14$
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Notes on Straight lines

Concepts Used:

Straight lines

A straight line is a line having the shortest distance between two points. 

A straight line can be represented as an equation in various forms,  as show in the image below:

 

The following are the many forms of the equation of the line that are presented in straight line-

1. Slope – Point Form

Assume P0(x0, y0) is a fixed point on a non-vertical line L with m as its slope. If P (x, y) is an arbitrary point on L, then the point (x, y) lies on the line with slope m through the fixed point (x0, y0) if and only if its coordinates fulfil the equation below.

y – y0 = m (x – x0)

2. Two – Point Form

Let's look at the line. L crosses between two places. P1(x1, y1) and P2(x2, y2)  are general points on L, while P (x, y) is a general point on L. As a result, the three points P1, P2, and P are collinear, and it becomes

The slope of P2P = The slope of P1P2 , i.e.

\(\frac{y-y_1}{x-x_1} = \frac{y_2-y_1}{x_2-x_1}\)

Hence, the equation becomes:

y - y1 =\( \frac{y_2-y_1}{x_2-x_1} (x-x1)\)

3. Slope-Intercept Form

Assume that a line L with slope m intersects the y-axis at a distance c from the origin, and that the distance c is referred to as the line L's y-intercept. As a result, the coordinates of the spot on the y-axis where the line intersects are (0, c). As a result, the slope of the line L is m, and it passes through a fixed point (0, c). The equation of the line L thus obtained from the slope – point form is given by

y – c =m( x - 0 )

As a result, the point (x, y) on the line with slope m and y-intercept c lies on the line, if and only if

y = m x +c