A(1, -2, 1) and B(2, -1, 2) are the end point | vedclass

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

Question

(A) The equation of the line $AB$ passing through $A(1, -2, 1)$ and $B(2, -1, 2)$ is given by $\frac{x-1}{2-1} = \frac{y-(-2)}{-1-(-2)} = \frac{z-1}{2

Vedclass · Accepted Answer

$18$

Vedclass · Answer

(A) The equation of the line $AB$ passing through $A(1, -2, 1)$ and $B(2, -1, 2)$ is given by $\frac{x-1}{2-1} = \frac{y-(-2)}{-1-(-2)} = \frac{z-1}{2-1}$,which simplifies to $\frac{x-1}{1} = \frac{y+2}{1} = \frac{z-1}{1} = K$.Any point $D$ on the line $AB$ can be represented as $(\alpha, \beta, \gamma) = (K+1, K-2, K+1)$.The direction ratios of the line $AB$ are $\langle 1, 1, 1 \rangle$.The vector $\vec{CD} = (\alpha-1, \beta-2, \gamma-3) = (K, K-4, K-2)$.Since $CD \perp AB$,the dot product of $\vec{CD}$ and the direction vector of $AB$ must be zero:$1(K) + 1(K-4) + 1(K-2) = 0$.$3K - 6 = 0 \Rightarrow K = 2$.Substituting $K=2$,we get $\alpha = 2+1 = 3$,$\beta = 2-2 = 0$,and $\gamma = 2+1 = 3$.Therefore,$\alpha^2 + \beta^2 + \gamma^2 = 3^2 + 0^2 + 3^2 = 9 + 0 + 9 = 18$.

$A(1, -2, 1)$ and $B(2, -1, 2)$ are the end points of a line segment. If $D(\alpha, \beta, \gamma)$ is the foot of the perpendicular drawn from $C(1, 2, 3)$ to $AB$,then $\alpha^2 + \beta^2 + \gamma^2 =$

Similar Questions

The angle between a line with direction ratios $2: 2: 1$ and a line joining the points $(3, 1, 4)$ and $(7, 2, 12)$ is

The point collinear with $(1, -2, -3)$ and $(2, 0, 0)$ among the following is

Find the equation of the line passing through the point $(1, 2, -4)$ and perpendicular to the two lines $\frac{x - 8}{3} = \frac{y + 19}{-16} = \frac{z - 10}{7}$ and $\frac{x - 15}{3} = \frac{y - 29}{8} = \frac{z - 5}{-5}$.

If the shortest distance between the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{\lambda}$ and $\frac{x-2}{1}=\frac{y-4}{4}=\frac{z-5}{5}$ is $\frac{1}{\sqrt{3}}$,then the sum of possible values of $\lambda$ is

The equation of a line passing through the point $(2, -1, 1)$ and parallel to the line whose equation is $\frac{x - 3}{2} = \frac{y + 1}{7} = \frac{z - 2}{-3}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module