The coordinates of the foot of the perpendicu | vedclass

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Question

(A) Let the line be $\frac{x+3}{5}=\frac{y+1}{2}=\frac{z+4}{3}=\lambda$. Any point $P$ on the line is given by $P(5\lambda-3, 2\lambda-1, 3\lambda-4)$

Vedclass · Accepted Answer

$\left(\frac{48}{19}, \frac{23}{19}, \frac{-13}{19}\right)$

Vedclass · Answer

(A) Let the line be $\frac{x+3}{5}=\frac{y+1}{2}=\frac{z+4}{3}=\lambda$. Any point $P$ on the line is given by $P(5\lambda-3, 2\lambda-1, 3\lambda-4)$. The given point is $A(0, 2, 3)$. The direction ratios of the line $AP$ are $(5\lambda-3-0, 2\lambda-1-2, 3\lambda-4-3)$,which simplifies to $(5\lambda-3, 2\lambda-3, 3\lambda-7)$. Since $AP$ is perpendicular to the given line with direction ratios $(5, 2, 3)$,their dot product must be zero: $5(5\lambda-3) + 2(2\lambda-3) + 3(3\lambda-7) = 0$. $25\lambda - 15 + 4\lambda - 6 + 9\lambda - 21 = 0$. $38\lambda - 42 = 0$. $\lambda = \frac{42}{38} = \frac{21}{19}$. Substituting $\lambda = \frac{21}{19}$ into the coordinates of $P$: $x = 5(\frac{21}{19}) - 3 = \frac{105-57}{19} = \frac{48}{19}$. $y = 2(\frac{21}{19}) - 1 = \frac{42-19}{19} = \frac{23}{19}$. $z = 3(\frac{21}{19}) - 4 = \frac{63-76}{19} = \frac{-13}{19}$. Thus,the foot of the perpendicular is $\left(\frac{48}{19}, \frac{23}{19}, \frac{-13}{19}\right)$.

The coordinates of the foot of the perpendicular from the point $(0,2,3)$ on the line $\frac{x+3}{5}=\frac{y+1}{2}=\frac{z+4}{3}$ are

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