The coordinates of the point where the line f | vedclass

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(A) Let the given line be $\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+3}{4}=\lambda$. Any point $P$ on the line is given by $P = (2\lambda+1, -3\lambda+2, 4

Vedclass · Accepted Answer

$(3, -1, 1)$

Vedclass · Answer

(A) Let the given line be $\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+3}{4}=\lambda$. Any point $P$ on the line is given by $P = (2\lambda+1, -3\lambda+2, 4\lambda-3)$. Since this point $P$ lies on the plane $2x+4y-z=1$,we substitute the coordinates of $P$ into the plane equation: $2(2\lambda+1) + 4(-3\lambda+2) - (4\lambda-3) = 1$. Expanding the terms: $4\lambda + 2 - 12\lambda + 8 - 4\lambda + 3 = 1$. Combining the $\lambda$ terms and constants: $-12\lambda + 13 = 1$. $-12\lambda = -12$,which gives $\lambda = 1$. Substituting $\lambda = 1$ back into the coordinates of $P$: $x = 2(1)+1 = 3$,$y = -3(1)+2 = -1$,$z = 4(1)-3 = 1$. Thus,the point of intersection is $(3, -1, 1)$.

The coordinates of the point where the line $\frac{x-1}{2}=\frac{y-2}{-3}=\frac{z+3}{4}$ meets the plane $2x+4y-z=1$ are

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