The coordinates of the foot of the perpendicu | vedclass

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(A) The given line is $\frac{6 - x}{-3} = \frac{y - 7}{2} = \frac{7 - z}{2}$.Rewriting it in standard form: $\frac{x - 6}{3} = \frac{y - 7}{2} = \frac

Vedclass · Accepted Answer

$(3, 5, 9)$

Vedclass · Answer

(A) The given line is $\frac{6 - x}{-3} = \frac{y - 7}{2} = \frac{7 - z}{2}$.Rewriting it in standard form: $\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{-2} = k$.Any point $M$ on the line is given by $(3k + 6, 2k + 7, -2k + 7)$.Let $A = (1, 2, 3)$. The vector $\vec{AM} = (3k + 6 - 1, 2k + 7 - 2, -2k + 7 - 3) = (3k + 5, 2k + 5, -2k + 4)$.The direction vector of the line is $\vec{v} = (3, 2, -2)$.Since $\vec{AM} \perp \vec{v}$,their dot product is zero: $\vec{AM} \cdot \vec{v} = 0$.$3(3k + 5) + 2(2k + 5) - 2(-2k + 4) = 0$.$9k + 15 + 4k + 10 + 4k - 8 = 0$.$17k + 17 = 0 \Rightarrow k = -1$.Substituting $k = -1$ into the coordinates of $M$: $x = 3(-1) + 6 = 3$,$y = 2(-1) + 7 = 5$,$z = -2(-1) + 7 = 9$.Thus,the coordinates of the foot of the perpendicular are $(3, 5, 9)$.

The coordinates of the foot of the perpendicular drawn from the point $(1, 2, 3)$ to the line $\frac{6 - x}{-3} = \frac{y - 7}{2} = \frac{7 - z}{2}$ are:

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