Question

Find the distance of the point P(3, 4, 4) from the point, where the line joining the points A(3, –4, –5) and B(2, –3, 1) intersects the plane 2x + y + z = 7.

Answer 1

The equation of the line passing through the points A(3, –4, –5) and B(2, –3, 1) is given by 

\[\frac{x - 3}{2 - 3} = \frac{y - \left( - 4 \right)}{- 3 - \left( - 4 \right)} = \frac{z - \left( - 5 \right)}{1 - \left( - 5 \right)}\]
\[\text{ Or }  \frac{x - 3}{- 1} = \frac{y + 4}{1} = \frac{z + 5}{6}\]

The coordinates of any point on the line 

\[\frac{x - 3}{- 1} = \frac{y + 4}{1} = \frac{z + 5}{6} = \lambda\left( \text{say}    \right)\] are \[\left( - \lambda + 3, \lambda - 4, 6\lambda - 5 \right)\]   ...(1)

If it lies on the plane 2x + y + x = 7, then 

\[2\left( - \lambda + 3 \right) + \left( \lambda - 4 \right) + \left( 6\lambda - 5 \right) = 7\]

\[ \Rightarrow 5\lambda - 3 = 7\]

\[ \Rightarrow 5\lambda = 10\]

\[ \Rightarrow \lambda = 2\] 

Putting \[\lambda = 2\]  in (1), we get (1, –2, 7) as the coordinates of the point of intersection of the given line and plane.

∴ Required distance = Distance between points (3, 4, 4) and (1, –2, 7)

\[= \sqrt{\left( 3 - 1 \right)^2 + \left( 4 + 2 \right)^2 + \left( 4 - 7 \right)^2}\]

\[ = \sqrt{4 + 36 + 9}\]

\[ = \sqrt{49}\]

\[ = 7 \text{ units} \]

Answer 2

The equation of the line passing through A(3, –4, –5) and B(2, –3, 1) is given by

`(x - 3)/(2 - 3) = (y + 4)/(-3 + 4) = (z + 5)/(1 + 5) = λ`

i.e., `(x - 3)/(-1) = (y + 4)/1 = (z + 5)/6 = λ`

Then co-ordinates of any random point on the line AB is Q(– λ + 3, λ – 4, 6λ – 5)

Line AB interests the plane 2λ + y + z = 7

Then 2(–λ + 3) + (λ – 4) + (6λ – 5) = 7

⇒ –2λ + 6 + λ – 4 + 6λ – 5 = 7

⇒ 5λ – 3 = 7

⇒ 5λ = 10

⇒ λ = 2

Therefore, co-ordinates of the point of interaction of the given line and the plane are Q(1, –2, 7)

Now, the distance between P(3, 4, 4) and Q(1, –2, 7)

= `sqrt((3 - 1)^2 + (4 + 2)^2 + (4 - 7)^2`

= `sqrt(4 + 36 + 9)`

= `sqrt(49)`

∴ PQ = 7 units