Let a triangle PQR be such that P and Q lie o | vedclass

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

Question

(C) Let the line be $L$. Any point on $L$ is given by $(8k-3, 2k+4, 2k-1)$.Since $P$ and $Q$ lie on $L$ and are at a distance of $6$ units from $R(1,

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(C) Let the line be $L$. Any point on $L$ is given by $(8k-3, 2k+4, 2k-1)$.Since $P$ and $Q$ lie on $L$ and are at a distance of $6$ units from $R(1, 2, 3)$, we use the distance formula: $(8k-3-1)^2 + (2k+4-2)^2 + (2k-1-3)^2 = 6^2$.This simplifies to $(8k-4)^2 + (2k+2)^2 + (2k-4)^2 = 36$.Expanding the squares: $(64k^2 - 64k + 16) + (4k^2 + 8k + 4) + (4k^2 - 16k + 16) = 36$.Combining like terms: $72k^2 - 72k + 36 = 36$, which gives $72k^2 - 72k = 0$.Factoring gives $72k(k-1) = 0$, so $k=0$ or $k=1$.For $k=0$, $P = (-3, 4, -1)$. For $k=1$, $Q = (5, 6, 1)$.The centroid $(\alpha, \beta, \gamma)$ is $(\frac{-3+5+1}{3}, \frac{4+6+2}{3}, \frac{-1+1+3}{3}) = (1, 4, 1)$.Thus, $\alpha + \beta + \gamma = 1 + 4 + 1 = 6$.

Let a triangle $PQR$ be such that $P$ and $Q$ lie on the line $\frac{x+3}{8} = \frac{y-4}{2} = \frac{z+1}{2}$ and are at a distance of $6$ units from $R(1, 2, 3)$. If $(\alpha, \beta, \gamma)$ is the centroid of $\triangle PQR$, then $\alpha + \beta + \gamma$ is equal to :

Similar Questions

If $\vec{r}=\hat{i}+\hat{j}+t(2 \hat{i}-\hat{j}+\hat{k})$ and $\vec{r}=2 \hat{i}-\hat{j}-\hat{k}+s(3 \hat{i}-5 \hat{j}+2 \hat{k})$ are the vector equations of two lines $L_1$ and $L_2$, then the shortest distance between them is

The angle between the lines whose direction cosines satisfy the equations $l+m+n=0$ and $l^2+m^2-n^2=0$ is

If lines $\frac{1-x}{3}=\frac{7y-14}{2p}=\frac{z-3}{2}$ and $\frac{7-7x}{3p}=\frac{y-5}{1}=\frac{6-z}{5}$ are mutually perpendicular to each other,then $p = $ . . . . . . .

The line $l_1$ passes through the point $(2, 6, 2)$ and is perpendicular to the plane $2x + y - 2z = 10$. Then the shortest distance between the line $l_1$ and the line $\frac{x + 1}{2} = \frac{y + 4}{-3} = \frac{z}{2}$ is :

The equation of the line passing through the point $(1, -3, 5)$ and making equal angles with the coordinate axes is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module