The point of intersection of the two lines fr | vedclass

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

Question

(D) For the first line: $\frac{x - 3}{3} = \frac{y - 3}{-1} = k$ and $z = 1$. Thus, $x = 3k + 3, y = -k + 3, z = 1$.For the second line: $\frac{x - 6}

Vedclass · Accepted Answer

$(6, 2, 1)$

Vedclass · Answer

(D) For the first line: $\frac{x - 3}{3} = \frac{y - 3}{-1} = k$ and $z = 1$. Thus, $x = 3k + 3, y = -k + 3, z = 1$.For the second line: $\frac{x - 6}{2} = \frac{z - 1}{3} = m$ and $y = 2$. Thus, $x = 2m + 6, y = 2, z = 3m + 1$.Equating the coordinates for intersection: $y = -k + 3 = 2 \implies k = 1$. Substituting $k=1$ into the first line gives $(x, y, z) = (3(1)+3, -1+3, 1) = (6, 2, 1)$.Checking the second line: $y = 2$ is satisfied. For $x = 6, z = 1$: $6 = 2m + 6 \implies m = 0$ and $1 = 3(0) + 1 \implies 1 = 1$. Both are satisfied.Thus, the point of intersection is $(6, 2, 1)$.

The point of intersection of the two lines $\frac{x - 3}{3} = \frac{y - 3}{-1}, z - 1 = 0$ and $\frac{x - 6}{2} = \frac{z - 1}{3}, y - 2 = 0$ is

Similar Questions

If the harmonic conjugate of $P(2, 3, 4)$ with respect to the line segment joining the points $A(3, -2, 2)$ and $B(6, -17, -4)$ is $Q(\alpha, \beta, \gamma)$, then $\alpha + \beta + \gamma =$

The foot of the perpendicular from $(0,2,3)$ to the line $\frac{x+3}{5}=\frac{y-1}{2}=\frac{z+4}{3}$ is

If the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-\lambda}{2}=\frac{z}{1}$ intersect each other,then $\lambda = \ldots$

$A(2,3,4), B(4,5,7), C(2,-6,3), D(4,-4, k)$ are four points. If the line $\overline{AB}$ is parallel to $\overline{CD}$,then $k$ is equal to

Show that the points $A(1, 2, 7)$,$B(2, 6, 3)$,and $C(3, 10, -1)$ are collinear.

Vedclass Test Series

Exam Paper Generator

Online Exam Module