Find the perpendicular distance from A1- 0- 3 to the line r-4- 7- 1-k1- 2- -2- k - R- Also find the foot of the perpendicular-

Consider #160; r⃗ = (4î + 7ĵ + k̂) + k(1î + 2ĵ 2k̂) Cartesian equation of the line #160; x 4/1=y 7/2=z 1/ 2 Let, #160; x 4/1=y 7/2=z 1/ 2=r ...

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Standard XII

Mathematics

Equation of Circle with (h,k) as Center

Find the perp...

Question

Find the perpendicular distance from $A (1, 0, 3)$ to the line $\to r = (4, 7, 1) + k (1, 2, - 2), k \in R$ . Also find the foot of the perpendicular.

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Solution

Consider

$\to r = (4^i + 7^j +^k) + k (1^i + 2^j - 2^k)$

Cartesian equation of the line

$\frac{x - 4}{1} = \frac{y - 7}{2} = \frac{z - 1}{- 2}$

Let,

$\frac{x - 4}{1} = \frac{y - 7}{2} = \frac{z - 1}{- 2} = r$

Then,

$x = r + 4, y = 2 r + 7, z = - 2 r + 1$

General point on the given line is
$(r + 4, 2 r + 7, - 2 r + 1)$

Let the foot of perpendicular to be,

$P (r_{1} + 4, 2 r_{1} + 7, - 2 r_{1} + 1)$

Now,
Direction ratios of the line $A P$ are

$r_{1} + 4 - 1, 2 r_{1} + 7 - 0, - 2 r_{1} + 1 - 3$
that is

$r_{1} + 3, 2 r_{1} + 7, - 2 r_{2} - 2$

Since $A P$ is perpendicular to the given line

Therefore,

$\begin{matrix} 1 (r_{1} + 3) + 2 (2 r_{1} + 7) - 2 (- 2 r_{1} - 2) = 0 r_{1} = - \frac{7}{3} \end{matrix}$

Then,

$\begin{matrix} r_{1} + 4 = \frac{5}{3} 2 r_{1} + 7 = \frac{7}{3} - 2 r_{1} + 1 = \frac{17}{3} \end{matrix}$

Therefore, foot of perpendicular is $P (\frac{5}{3}, \frac{7}{3}, \frac{17}{3})$

And perpendicular distance $= A P$

$\begin{matrix} = \sqrt{{(\frac{5}{3} - 1)}^{2} + {(\frac{7}{3})}^{2} + {(\frac{17}{3} - 3)}^{2}} = \sqrt{{(\frac{2}{3})}^{2} + {(\frac{7}{3})}^{2} + {(\frac{8}{3})}^{2}} = \frac{\sqrt{4 + 49 + 64}}{3} = \frac{\sqrt{117}}{3} \end{matrix}$

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Find the perpendicular distance from A(1,0,3) to the line →r=(4,7,1)+k(1,2,−2),k∈R. Also find the foot of the perpendicular.

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