Name: Equation of a Line in 3D (Symmetric Form)
Uploaded: 2019-07-19
Duration: 11 min 8 s
Description: We have seen the equation of a line in 2D. Here, in this video, we will see the equation of a line in 3D and also see why vectors play an important role in 3...

Question

Show that the foot of the perpendicular from the origin to the join of

A (- 9, 4, 5)

and

B (11, 0, - 1)

is the mid-point of

A B

. Also find distance of point

(2, 4, 4)

from the line

A B

.

Accepted Answer

Given points are $A (- 9, 4, 5) & B (11, 0, - 1)$ .

As the direction ratios of line joining points $(x_{1}, y_{1}) & (x_{2}, y_{2}) = x_{2} - x_{1}, y_{2} - y_{1}, z_{2} - z_{1}$ .

$∴$ Direction ratios of line $A B = 20, - 4, - 6$ .

So, equation of line $A B$ is given by,

$\frac{x + 9}{20} = \frac{y - 4}{- 4} = \frac{z - 5}{- 6} = λ$ (say),

where $(20, - 4, - 6) or (10, - 2, - 3)$ are the D.R.'s of line $A B$ .

Let $Q$ be the foot of perpendicular from origin to the line $A B$ .

So, the coordinates of $Q$ be $(20 λ - 9, - 4 λ + 4, - 6 λ + 5)$ .

And Direction ratios of $O Q = (20 λ - 9, - 4 λ + 4, - 6 λ + 5)$ .

Since $O Q$ is perpendicular to line $A B$ ,

$∴$ Sum of product of their corresponding Direction ratios is zero.

$\Rightarrow 10 (20 λ - 9) - 2 (- 4 λ + 4) - 3 (- 6 λ + 5) = 0$

$\Rightarrow 200 λ - 90 + 8 λ - 8 + 18 λ - 15 = 0$

$\Rightarrow 226 λ = 113$

$\Rightarrow λ = \frac{113}{226} \Rightarrow λ = \frac{1}{2}$

$∴$ Coordinates of $Q$ are $(20 \cdot \frac{1}{2} - 9, - 4 \cdot \frac{1}{2} + 4, - 6 \cdot \frac{1}{2} + 5)$ .

$= (1, 2, 2)$

And the mid-point of line joining $A (- 9, 4, 5) & B (11, 0, - 1)$ is

$(\frac{- 9 + 11}{2}, \frac{4 + 0}{2}, \frac{5 - 1}{2}) = (1, 2, 2) = Q$ .

Let $P$ be the point $(2, 4, 4) & Q (20 λ - 9, - 4 λ + 4, - 6 λ + 5)$ be any point on line $A B$ such that line $P Q^{'}$ is perpendicular to line $A B$ .

Now, Direction ratios of $P Q^{'}$ are $(20 λ - 9 - 2, - 4 + 4 - 4, - 6 λ + 5 - 4)$

$= (20 λ - 11, - 4 λ, - 6 λ + 1)$

$∴$ Sum of product of D.R.'s of $P Q^{'} & A B$ is zero.

$\Rightarrow 10 (20 λ - 11) - 2 (- 4 λ) - 3 (- 6 λ + 1) = 0$

$\Rightarrow 200 λ - 110 + 8 λ + 18 λ - 3 = 0$

$\Rightarrow 226 λ = 113 \Rightarrow λ = \frac{113}{226} = λ = \frac{1}{2}$

So coordinates of $Q^{'}$ are $(1, 2, 2)$ .

$∴$ Distance between $(2, 4, 4) & (1, 2, 2)$ is given by,

$= \sqrt{{(2 - 1)}^{2} + {(4 - 2)}^{2} + {(4 - 2)}^{2}}$ .

$= \sqrt{1^{2} + 2^{2} + 2^{2}} = \sqrt{9} = 3$

Show that the foot of the perpendicular from the origin to the join of $A (- 9, 4, 5)$ and $B (11, 0, - 1)$ is the mid-point of $A B$ . Also find distance of point $(2, 4, 4)$ from the line $A B$ .

Important Questions on Three Dimensional Geometry

Learn and Prepare for any exam you want !

Show that the foot of the perpendicular from the origin to the join of A-9,4,5 and B11,0,-1 is the mid-point of AB. Also find distance of point 2,4,4 from the line AB.

Important Questions on Three Dimensional Geometry

Learn and Prepare for any exam you want !

Show that the foot of the perpendicular from the origin to the join of $A (- 9, 4, 5)$ and $B (11, 0, - 1)$ is the mid-point of $A B$ . Also find distance of point $(2, 4, 4)$ from the line $A B$ .