The length of the projection of the line segm | vedclass

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

Question

(A) Let the points be $A(3, 4, 5)$,$B(4, 6, 3)$,$C(-1, 2, 4)$,and $D(1, 0, 5)$.First,we find the vector $\overrightarrow{AB}$ representing the line se

Vedclass · Accepted Answer

$\frac{4}{3}$

Vedclass · Answer

(A) Let the points be $A(3, 4, 5)$,$B(4, 6, 3)$,$C(-1, 2, 4)$,and $D(1, 0, 5)$.First,we find the vector $\overrightarrow{AB}$ representing the line segment joining $A$ and $B$:$\overrightarrow{AB} = (4-3)\hat{i} + (6-4)\hat{j} + (3-5)\hat{k} = \hat{i} + 2\hat{j} - 2\hat{k}$.Next,we find the vector $\overrightarrow{CD}$ representing the line joining $C$ and $D$:$\overrightarrow{CD} = (1 - (-1))\hat{i} + (0-2)\hat{j} + (5-4)\hat{k} = 2\hat{i} - 2\hat{j} + \hat{k}$.The length of the projection of vector $\overrightarrow{AB}$ on vector $\overrightarrow{CD}$ is given by the formula $\left| \frac{\overrightarrow{AB} \cdot \overrightarrow{CD}}{|\overrightarrow{CD}|} \right|$.Calculate the dot product $\overrightarrow{AB} \cdot \overrightarrow{CD} = (1)(2) + (2)(-2) + (-2)(1) = 2 - 4 - 2 = -4$.Calculate the magnitude of $\overrightarrow{CD} = |\overrightarrow{CD}| = \sqrt{2^2 + (-2)^2 + 1^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3$.Therefore,the length of the projection is $\left| \frac{-4}{3} \right| = \frac{4}{3}$.

The length of the projection of the line segment joining the points $(3, 4, 5)$ and $(4, 6, 3)$ on the line joining the points $(-1, 2, 4)$ and $(1, 0, 5)$ is

Similar Questions

Let the direction cosines of two lines satisfy the equations $3l+2m+n=0$ and $2mn-3nl+5lm=0$. If $\theta$ is the angle between these two lines,then $\cos \theta=$

If $\alpha, \beta$ and $\gamma$ are the angles which a half ray makes with the positive direction of the axes,then $\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma$ is equal to

$A$ line makes angles of $45^\circ$ and $60^\circ$ with the positive axes of $X$ and $Y$ respectively. The angle made by the same line with the positive axis of $Z$ is:

If $\theta$ is the acute angle between the two lines whose direction cosines are connected by the relations $l+m+n=0$ and $2lm+2nl-mn=0$,then $\cos \theta=$

If the direction cosines of a line satisfy the relations $l-m+n=0$ and $lm+mn-4nl=0$,then the direction cosines of the line are

Vedclass Test Series

Exam Paper Generator

Online Exam Module