The line frac{x + 1}{2} = frac{y + 1}{3} = fr | vedclass

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(A) Let the line be $\frac{x+1}{2}=\frac{y+1}{3}=\frac{z+1}{4}=\lambda$. Then,the coordinates of any point on the line are given by $x = 2\lambda - 1$

Vedclass · Accepted Answer

$(1, 2, 3)$

Vedclass · Answer

(A) Let the line be $\frac{x+1}{2}=\frac{y+1}{3}=\frac{z+1}{4}=\lambda$. Then,the coordinates of any point on the line are given by $x = 2\lambda - 1$,$y = 3\lambda - 1$,and $z = 4\lambda - 1$. Since this point lies on the plane $x + 2y + 3z = 14$,we substitute these values into the plane equation: $(2\lambda - 1) + 2(3\lambda - 1) + 3(4\lambda - 1) = 14$. Expanding the equation: $2\lambda - 1 + 6\lambda - 2 + 12\lambda - 3 = 14$. Combining the terms: $20\lambda - 6 = 14$. $20\lambda = 20$,which gives $\lambda = 1$. Substituting $\lambda = 1$ back into the coordinate expressions: $x = 2(1) - 1 = 1$,$y = 3(1) - 1 = 2$,$z = 4(1) - 1 = 3$. Thus,the point of intersection is $(1, 2, 3)$.

The line $\frac{x + 1}{2} = \frac{y + 1}{3} = \frac{z + 1}{4}$ meets the plane $x + 2y + 3z = 14$ at the point:

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