Statement-1: The distance between the two par | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) For Statement-$2$: By definition,the distance between two parallel lines is the perpendicular distance from any point on one line to the other lin

Vedclass · Accepted Answer

Statement-$1$ is true,Statement-$2$ is true. Statement-$2$ is the correct explanation for Statement-$1$.

Vedclass · Answer

(A) For Statement-$2$: By definition,the distance between two parallel lines is the perpendicular distance from any point on one line to the other line. Thus,Statement-$2$ is true.For Statement-$1$: Let the lines be $L_1: \frac{x}{2} = \frac{y}{-1} = \frac{z}{2}$ and $L_2: \frac{x-1}{4} = \frac{y-1}{-2} = \frac{z-1}{4}$.Point $P(0, 0, 0)$ lies on $L_1$. The direction vector of $L_2$ is $\vec{b} = 4\hat{i} - 2\hat{j} + 4\hat{k}$. $A$ point on $L_2$ is $Q(1, 1, 1)$.The vector $\vec{PQ} = (1-0)\hat{i} + (1-0)\hat{j} + (1-0)\hat{k} = \hat{i} + \hat{j} + \hat{k}$.The distance $d$ between parallel lines is given by $d = \frac{|\vec{PQ} 	imes \vec{b}|}{|\vec{b}|}$.$\vec{PQ} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & 1 \ 4 & -2 & 4 \end{vmatrix} = \hat{i}(4 - (-2)) - \hat{j}(4 - 4) + \hat{k}(-2 - 4) = 6\hat{i} + 0\hat{j} - 6\hat{k}$.$|\vec{PQ} 	imes \vec{b}| = \sqrt{6^2 + 0^2 + (-6)^2} = \sqrt{36 + 36} = \sqrt{72} = 6\sqrt{2}$.$|\vec{b}| = \sqrt{4^2 + (-2)^2 + 4^2} = \sqrt{16 + 4 + 16} = \sqrt{36} = 6$.$d = \frac{6\sqrt{2}}{6} = \sqrt{2}$.Thus,Statement-$1$ is true and Statement-$2$ is the correct explanation.

Similar Questions

Find the distance of a point $(2, 4, -1)$ from the line $\frac{x+5}{1} = \frac{y+3}{4} = \frac{z-6}{-9}$.

Let $L$ be the line parallel to the vector $\sqrt{2} \hat{i}-5 \hat{j}+3 \hat{k}$ and passing through the point $A$ given by $\hat{i}+2 \hat{j}-3 \hat{k}$. If the distance between $A$ and a point $P$ on the line $L$ is $18$ units, then the position vector of such a point $P$ is

If the direction ratios of two lines are given by $3lm - 4ln + mn = 0$ and $l + 2m + 3n = 0$,then the angle between the lines is

$A$ line with direction cosines proportional to $2, 1, 2$ meets the line $L_1$ passing through $(0, -1, 0)$ with direction ratios $1, 1, 1$ at $A(x, y, z)$ and another line $L_2$ at $B(1, 1, 1)$. Then $x+y+z=$

The square of the distance of the point $\left(\frac{15}{7}, \frac{32}{7}, 7\right)$ from the line $\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}$ in the direction of the vector $\hat{i}+4 \hat{j}+7 \hat{k}$ is :

Vedclass Test Series

Exam Paper Generator

Online Exam Module