A line passes through A(4, -6, -2) and B(16, | vedclass

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

Question

(D) The direction vector of line $AB$ is $\vec{v} = (16-4, -2-(-6), 4-(-2)) = (12, 4, 6)$.The unit vector in the direction of $AB$ is $\hat{u} = \frac

Vedclass · Accepted Answer

$22$

Vedclass · Answer

(D) The direction vector of line $AB$ is $\vec{v} = (16-4, -2-(-6), 4-(-2)) = (12, 4, 6)$.The unit vector in the direction of $AB$ is $\hat{u} = \frac{(12, 4, 6)}{\sqrt{12^2 + 4^2 + 6^2}} = \frac{(12, 4, 6)}{\sqrt{144 + 16 + 36}} = \frac{(12, 4, 6)}{\sqrt{196}} = \frac{(12, 4, 6)}{14} = (\frac{6}{7}, \frac{2}{7}, \frac{3}{7})$.Point $P$ is at a distance of $21$ units from $A(4, -6, -2)$ along the line $AB$,so $P = A + 21 \hat{u}$.$P = (4, -6, -2) + 21(\frac{6}{7}, \frac{2}{7}, \frac{3}{7}) = (4 + 18, -6 + 6, -2 + 9) = (22, 0, 7)$.Here $a=22, b=0, c=7$,which are non-negative integers.The distance between $P(22, 0, 7)$ and $Q(4, -12, 3)$ is $\sqrt{(22-4)^2 + (0-(-12))^2 + (7-3)^2}$.$= \sqrt{18^2 + 12^2 + 4^2} = \sqrt{324 + 144 + 16} = \sqrt{484} = 22$.

$A$ line passes through $A(4, -6, -2)$ and $B(16, -2, 4)$. The point $P(a, b, c)$,where $a, b, c$ are non-negative integers,lies on the line $AB$ at a distance of $21$ units from point $A$. The distance between the points $P(a, b, c)$ and $Q(4, -12, 3)$ is equal to...........

Similar Questions

The shortest distance between the skew lines $r = (-\hat{i} + 3\hat{k}) + t(2\hat{i} + 3\hat{j} + 6\hat{k})$ and $r = (3\hat{i} + \hat{j} - \hat{k}) + s(2\hat{i} - \hat{j} + 2\hat{k})$ is

Let the line $L$ intersect the lines $x-2=-y=z-1$ and $2(x+1)=2(y-1)=z+1$,and be parallel to the line $\frac{x-2}{3}=\frac{y-1}{1}=\frac{z-2}{2}$. Then which of the following points lies on $L$?

Line $L_1$ passes through the points $\hat{i}+\hat{j}$ and $\hat{k}-\hat{i}$. Line $L_2$ passes through the point $\hat{j}+2\hat{k}$ and is parallel to the vector $\hat{i}+\hat{j}+\hat{k}$. If $x\hat{i}+y\hat{j}+z\hat{k}$ is the point of intersection of the lines $L_1$ and $L_2$,then $(y-x)=$

If for some $\alpha \in R$,the lines $L_1: \frac{x+1}{2}=\frac{y-2}{-1}=\frac{z-1}{1}$ and $L_2: \frac{x+2}{\alpha}=\frac{y+1}{5-\alpha}=\frac{z+1}{1}$ are coplanar,then the line $L_2$ passes through the point

The angle between two lines $\frac{x + 1}{2} = \frac{y + 3}{2} = \frac{z - 4}{-1}$ and $\frac{x - 4}{1} = \frac{y + 4}{2} = \frac{z + 1}{2}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module