A line L passes through the points hat{i}+2 h | vedclass

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

Question

(B) The equation of the line $L$ passing through $(1, 2, 1)$ and $(-2, 0, 3)$ is given by $\frac{x-1}{-3} = \frac{y-2}{-2} = \frac{z-1}{2} = \lambda$.

Vedclass · Accepted Answer

$-8 \hat{i}-4 \hat{j}+7 \hat{k}$

Vedclass · Answer

(B) The equation of the line $L$ passing through $(1, 2, 1)$ and $(-2, 0, 3)$ is given by $\frac{x-1}{-3} = \frac{y-2}{-2} = \frac{z-1}{2} = \lambda$. Any point on this line is $A(-3\lambda + 1, -2\lambda + 2, 2\lambda + 1)$. The plane $P$ passes through $(0, 0, 0)$, $(0, 0, 4)$, and $(2, 1, 0)$. The equation of the plane is given by the determinant $\begin{vmatrix} x & y & z \ 0 & 0 & 4 \ 2 & 1 & 0 \end{vmatrix} = 0$. Expanding the determinant, we get $-4(x - 2y) = 0$, which simplifies to $x - 2y = 0$. Since point $A$ lies on the plane, substitute the coordinates of $A$ into the plane equation: $(-3\lambda + 1) - 2(-2\lambda + 2) = 0$. $-3\lambda + 1 + 4\lambda - 4 = 0 \Rightarrow \lambda - 3 = 0 \Rightarrow \lambda = 3$. Substituting $\lambda = 3$ into the coordinates of $A$: $x = -3(3) + 1 = -8$, $y = -2(3) + 2 = -4$, $z = 2(3) + 1 = 7$. Thus, the point is $-8\hat{i} - 4\hat{j} + 7\hat{k}$.

$A$ line $L$ passes through the points $\hat{i}+2 \hat{j}+\hat{k}$ and $-2 \hat{i}+3 \hat{k}$. $A$ plane $P$ passes through the origin and the points $4 \hat{k}, 2 \hat{i}+\hat{j}$. The point where the line $L$ meets the plane $P$ is

Similar Questions

The direction ratios of the line of intersection of the planes $x-y+z-5=0$ and $x-3y-6=0$ are:

If the lines $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z+3}{1}$ and $\frac{x-a}{2}=\frac{y+2}{3}=\frac{z-3}{1}$ intersect at the point $P$,then the distance of the point $P$ from the plane $z = a$ is:

The distance of the origin from the plane $r \cdot(3 \hat{i}+4 \hat{j}-12 \hat{k})=7$ measured parallel to the line $r=(\hat{i}+2 \hat{j}+3 \hat{k})+t(6 \hat{i}+2 \hat{j}+3 \hat{k})$ is

Let the line $\frac{x}{1}=\frac{6-y}{2}=\frac{z+8}{5}$ intersect the lines $\frac{x-5}{4}=\frac{y-7}{3}=\frac{z+2}{1}$ and $\frac{x+3}{6}=\frac{3-y}{3}=\frac{z-6}{1}$ at the points $A$ and $B$ respectively. Then the distance of the mid-point of the line segment $AB$ from the plane $2x-2y+z=14$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module