For non-coplanar vectors a, b and c, if the p | vedclass

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

Question

(C) The equation of the plane is given by $r = b + c + x(a - b) + y(c + a) = (x + y)a + (1 - x)b + (1 + y)c$ $\ldots(i)$.The equation of the line is g

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) The equation of the plane is given by $r = b + c + x(a - b) + y(c + a) = (x + y)a + (1 - x)b + (1 + y)c$ $\ldots(i)$.The equation of the line is given by $r = a + t(b - c) = a + tb - tc$ $\ldots(ii)$.Since the point of intersection satisfies both equations, we equate the coefficients of $a, b,$ and $c$ because $a, b, c$ are non-coplanar:$x + y = 1$ $\ldots(iii)$$1 - x = t$ $\ldots(iv)$$1 + y = -t$ $\ldots(v)$Adding equations $(iv)$ and $(v)$, we get $2 - x + y = 0$, so $x - y = 2$ $\ldots(vi)$.Adding $(iii)$ and $(vi)$, we get $2x = 3$, so $x = \frac{3}{2}$.Substituting $x = \frac{3}{2}$ into $(iii)$, we get $y = 1 - \frac{3}{2} = -\frac{1}{2}$.Substituting $x = \frac{3}{2}$ into $(iv)$, we get $t = 1 - \frac{3}{2} = -\frac{1}{2}$.Now, the point of intersection $r$ is $a + t(b - c) = a - \frac{1}{2}(b - c) = a - \frac{1}{2}b + \frac{1}{2}c$.Comparing this with $la + mb + nc$, we get $l = 1, m = -\frac{1}{2}, n = \frac{1}{2}$.Finally, $3l + 4m + 2n = 3(1) + 4(-\frac{1}{2}) + 2(\frac{1}{2}) = 3 - 2 + 1 = 2$.

For non-coplanar vectors $a, b$ and $c$, if the point of intersection of the line $r=a+t(b-c)$ and the plane $r=b+c+x(a-b)+y(c+a)$ is $l a+m b+n c$, then $3 l+4 m+2 n=$

Similar Questions

Find the equation of the line passing through the point $(3, 0, 1)$ and parallel to the planes $x+2y=0$ and $3y-z=0$.

If the distance of the point $2 \hat{i} + 3 \hat{j} + \lambda \hat{k}$ from the plane $\vec{r} \cdot (3 \hat{i} + 2 \hat{j} + 6 \hat{k}) = 13$ is $5$ units,then $\lambda =$

If the 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 equation of the plane containing these lines is:

If $4x + 4y - kz = 0$ is the equation of the plane through the origin that contains the line $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z}{4},$ then $k =$

The equation of the plane containing the straight line $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$ and perpendicular to the plane containing the straight lines $\frac{x}{3} = \frac{y}{4} = \frac{z}{2}$ and $\frac{x}{4} = \frac{y}{2} = \frac{z}{3}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module