The point of intersection of the line joining | vedclass

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

Question

(D) The line passes through $A(1, 2, 1)$ and $B(2, -1, -1)$. The direction vector is $\vec{v} = B - A = (2-1)\bar{i} + (-1-2)\bar{j} + (-1-1)\bar{k} =

Vedclass · Accepted Answer

$\frac{1}{7}(15\bar{i} - 10\bar{j} - 9\bar{k})$

Vedclass · Answer

(D) The line passes through $A(1, 2, 1)$ and $B(2, -1, -1)$. The direction vector is $\vec{v} = B - A = (2-1)\bar{i} + (-1-2)\bar{j} + (-1-1)\bar{k} = \bar{i} - 3\bar{j} - 2\bar{k}$. The equation of the line is $\vec{r} = (1 + t)\bar{i} + (2 - 3t)\bar{j} + (1 - 2t)\bar{k}$.The plane passes through $P(1, 0, 0)$, $Q(0, 2, 0)$, and $R(0, 0, 3)$. The intercept form of the plane is $\frac{x}{1} + \frac{y}{2} + \frac{z}{3} = 1$, which simplifies to $6x + 3y + 2z = 6$.Substituting the line coordinates into the plane equation: $6(1 + t) + 3(2 - 3t) + 2(1 - 2t) = 6$.$6 + 6t + 6 - 9t + 2 - 4t = 6$.$14 - 7t = 6 \implies 7t = 8 \implies t = \frac{8}{7}$.Substituting $t = \frac{8}{7}$ into the line equation: $x = 1 + \frac{8}{7} = \frac{15}{7}$, $y = 2 - 3(\frac{8}{7}) = \frac{14 - 24}{7} = -\frac{10}{7}$, $z = 1 - 2(\frac{8}{7}) = \frac{7 - 16}{7} = -\frac{9}{7}$.The intersection point is $\frac{1}{7}(15\bar{i} - 10\bar{j} - 9\bar{k})$.

The point of intersection of the line joining the points $\bar{i} + 2\bar{j} + \bar{k}$ and $2\bar{i} - \bar{j} - \bar{k}$ and the plane passing through the points $\bar{i}, 2\bar{j}, 3\bar{k}$ is:

Similar Questions

The equation of the plane passing through the intersection of the lines $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-5}{-3}$ and $\frac{x+5}{3}=\frac{y-4}{-1}=\frac{z+3}{4}$ and parallel to the $xy$-plane is

Find the coordinates of the point where the line joining the points $(2, -3, 1)$ and $(3, -4, -5)$ intersects the plane $2x + y + z = 7$.

Let the plane $P: 4x - y + z = 10$ be rotated by an angle $\frac{\pi}{2}$ about its line of intersection with the plane $x + y - z = 4$. If $\alpha$ is the distance of the point $(2, 3, -4)$ from the new position of the plane $P$,then $35\alpha$ is

If the straight lines $\frac{x-1}{2}=\frac{y+1}{k}=\frac{z}{2}$ and $\frac{x+1}{5}=\frac{y+1}{2}=\frac{z}{k}$ are coplanar,then the plane$(s)$ containing these two lines is(are):

The lines $\frac{x - a + d}{\alpha - \delta} = \frac{y - a}{\alpha} = \frac{z - a - d}{\alpha + \delta}$ and $\frac{x - b + c}{\beta - \gamma} = \frac{y - b}{\beta} = \frac{z - b - c}{\beta + \gamma}$ are coplanar. Find the equation of the plane in which they lie.

Vedclass Test Series

Exam Paper Generator

Online Exam Module