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Byju's Answer

Standard XII

Mathematics

Equation of a Line Passing through Two Given Points

Let Q be the ...

Question

Let $Q$ be the foot of the perpendicular from the point $P (7, - 2, 13)$ on the plane containing the lines $\frac{x + 1}{6} = \frac{y - 1}{7} = \frac{z - 3}{8}$ and $\frac{x - 1}{3} = \frac{y - 2}{5} = \frac{z - 3}{7} .$ Then $(P Q)^{2},$ is equal to

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Solution

Normal vector for plane:
$= ∣ ∣ ∣ ∣ ∣ \begin{matrix} ^i & ^j & ^k 6 & 7 & 8 3 & 5 & 7 \end{matrix} ∣ ∣ ∣ ∣ ∣$
$= 9^i - 18^j + 9^k = 9 (^i - 2^j +^k)$
$\Rightarrow$ Normal is parallel to $^i - 2^j +^k$
Plane passes through $(1, 2, 3)$ as it is a point on $L_{2}$ so equation of plane
$1 (x - 1) - 2 (y - 2) + 1 (z - 3) = 0$
$x - 2 y + z = 0$
$P Q = \frac{7 - 2 (- 2) + 13}{\sqrt{6}} \Rightarrow P Q^{2} = 96$

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Similar questions

Q. The coordinates of the foot of the perpendicular from the points

(1, - 2, 1)

on the plane containing the lines,

\frac{x + 1}{6} = \frac{y - 1}{7} = \frac{z - 3}{8}

and

\frac{x - 1}{3} = \frac{y - 2}{5} = \frac{z - 3}{7}

is:

Q. In

R^{3},

let

P_{1} : x - y + z = 3

and

L : \frac{x - 1}{3} = \frac{y - 2}{2} = \frac{z + 1}{1}

be the equation of plane and straight line respectively. Let

Q

be the image of the point

(- 1, 2, 3)

with respect to the plane

P_{1} .

Then the equation of the plane

P_{2}

passing through

Q

and containing the line

L

Perpendicular are drawn from points on the line $\frac{x + 2}{2} = \frac{y + 1}{- 1} = \frac{z}{3}$ to the plane x + y + z = 3. The feet of perpendiculars lie on the line

Q. The equation of the plane containing the two lines of intersection of the two pairs of planes x + 2y – z – 3 = 0 and 3x – y + 2z – 1 = 0, 2x – 2y + 3z = 0 and x – y + z + 1 =0 is :

Q. The perpendicular distance from the origin to the plane containing the two lines,

\frac{x + 2}{3} = \frac{y - 2}{5} = \frac{z + 5}{7}

and

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

, is:

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Let Q be the foot of the perpendicular from the point P(7,−2,13) on the plane containing the lines x+16=y−17=z−38 and x−13=y−25=z−37. Then (PQ)2, is equal to

Normal vector for plane: =∣∣ ∣ ∣∣^i^j^k678357∣∣ ∣ ∣∣ =9^i−18^j+9^k=9(^i−2^j+^k) ⇒ Normal is parallel to ^i−2^j+^k Plane passes through (1,2,3) as it is a point on L2 so equation of plane 1(x−1)−2(y−2)+1(z−3)=0 x−2y+z=0 PQ=7−2(−2)+13√6⇒PQ2=96

Let $Q$ be the foot of the perpendicular from the point $P (7, - 2, 13)$ on the plane containing the lines $\frac{x + 1}{6} = \frac{y - 1}{7} = \frac{z - 3}{8}$ and $\frac{x - 1}{3} = \frac{y - 2}{5} = \frac{z - 3}{7} .$ Then $(P Q)^{2},$ is equal to