If A(1, 0, 2),B(2, 1, 0),C(2, -5, 3),and D(0, | vedclass

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

Question

(A) The equation of line $AB$ passing through $A(1, 0, 2)$ and $B(2, 1, 0)$ is $\frac{x-1}{2-1} = \frac{y-0}{1-0} = \frac{z-2}{0-2} = \lambda$ $\Right

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(A) The equation of line $AB$ passing through $A(1, 0, 2)$ and $B(2, 1, 0)$ is $\frac{x-1}{2-1} = \frac{y-0}{1-0} = \frac{z-2}{0-2} = \lambda$ $\Rightarrow \frac{x-1}{1} = \frac{y}{1} = \frac{z-2}{-2} = \lambda$. Any point on $AB$ is $P(\lambda+1, \lambda, -2\lambda+2)$. The equation of line $CD$ passing through $C(2, -5, 3)$ and $D(0, 3, 2)$ is $\frac{x-2}{0-2} = \frac{y+5}{3+5} = \frac{z-3}{2-3} = \mu$ $\Rightarrow \frac{x-2}{-2} = \frac{y+5}{8} = \frac{z-3}{-1} = \mu$. Any point on $CD$ is $Q(-2\mu+2, 8\mu-5, -\mu+3)$. For intersection,$P = Q$: $\lambda+1 = -2\mu+2 \Rightarrow \lambda + 2\mu = 1$ $\lambda = 8\mu-5 \Rightarrow \lambda - 8\mu = -5$ Subtracting the equations: $10\mu = 6 \Rightarrow \mu = \frac{3}{5}$. Substituting $\mu$: $\lambda = 8(\frac{3}{5}) - 5 = \frac{24-25}{5} = -\frac{1}{5}$. Wait,checking intersection: $\lambda+1 = -2(3/5)+2 = 4/5$,$\lambda = -1/5$. Re-calculating: $\lambda+2\mu=1$ and $\lambda-8\mu=-5$. Subtracting: $10\mu=6 \Rightarrow \mu=0.6$. $\lambda = 1 - 1.2 = -0.2$. Using $\lambda = -0.2$: $x = 0.8, y = -0.2, z = 2.4$. Check $CD$: $x = -2(0.6)+2 = 0.8, y = 8(0.6)-5 = -0.2, z = -0.6+3 = 2.4$. $P(0.8, -0.2, 2.4) = (4/5, -1/5, 12/5)$. $a+b+c = (4-1+12)/5 = 15/5 = 3$.

If $A(1, 0, 2)$,$B(2, 1, 0)$,$C(2, -5, 3)$,and $D(0, 3, 2)$ are four points and the point of intersection of the lines $AB$ and $CD$ is $P(a, b, c)$,then $a + b + c =$

Similar Questions

If the Cartesian equation of the line is $x-1=2y+3=3-z$,then its vector equation is

The direction cosines of the line which is perpendicular to the lines $\frac{x-7}{2}=\frac{y+17}{-3}=\frac{z-6}{1}$ and $\frac{x+5}{1}=\frac{y+3}{2}=\frac{z-6}{-2}$ are

The equation of the line passing through $(1, 2, 3)$ and perpendicular to the lines $x-1 = \frac{y+2}{2} = \frac{z+4}{4}$ and $\frac{x-1}{2} = \frac{y-2}{2} = z+3$ is

The equation of the line passing through $(2, 3, 4)$ and parallel to the $Y$-axis is . . . . . . .

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module