Find the coordinates of the point of intersec | vedclass

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

Question

(D) Let the general point on the line be $P(6 - \lambda, -1, -3 + 4\lambda)$.Since this point lies on the plane $x + y - z = 3$,we substitute the coor

Vedclass · Accepted Answer

$(5, -1, 1)$

Vedclass · Answer

(D) Let the general point on the line be $P(6 - \lambda, -1, -3 + 4\lambda)$.Since this point lies on the plane $x + y - z = 3$,we substitute the coordinates into the plane equation:$(6 - \lambda) + (-1) - (-3 + 4\lambda) = 3$$6 - \lambda - 1 + 3 - 4\lambda = 3$$8 - 5\lambda = 3$$5\lambda = 5$$\lambda = 1$Substituting $\lambda = 1$ back into the point $P$,we get:$x = 6 - 1 = 5$$y = -1$$z = -3 + 4(1) = 1$Thus,the point of intersection is $(5, -1, 1)$.

Find the coordinates of the point of intersection of the line $\frac{x - 6}{-1} = \frac{y + 1}{0} = \frac{z + 3}{4}$ and the plane $x + y - z = 3$.

Similar Questions

The plane containing the line $\frac{x - 3}{2} = \frac{y + 2}{-1} = \frac{z - 1}{3}$ and also containing its projection on the plane $2x + 3y - z = 5$ contains which one of the following points?

Let the image of the point $P(2, -1, 3)$ in the plane $x + 2y - z = 0$ be $Q$. Then the distance of the plane $3x + 2y + z + 29 = 0$ from the point $Q$ is $.........$.

Let the line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ lie in the plane $x+3y-\alpha z+\beta=0$. Then the value of $(\beta-\alpha)$ is equal to

The distance of the point $O(\vec{0})$ from the plane $\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=5$ measured parallel to the vector $2 \hat{i}+3 \hat{j}-6 \hat{k}$ is:

The line of intersection of the planes $x + 2y = 0$ and $y - 3z + 3 = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module