Question

Find the equation of a plane which bisects perpendicularly the line joining the points A(2,3,4) and B(4,5,8) at right angles.​

Answer 1

According to question,

The equation of a plane is bisecting perpendicularly the line joining the points A(2,3,4) and B(4,5,8) at right angles.

So,

Mid points of A = $\bigg( \dfrac{ x_{1} \:+\:x_{2}}{2} \bigg)$, $\bigg( \dfrac{ y_{1} \:+\:y_{2}}{2} \bigg)$, $\bigg( \dfrac{ z_{1} \:+\:z_{2}}{2} \bigg)$,

=> $\dfrac{2\:+\:4}{2}$, $\dfrac{3\:+\:5}{2}$, $\dfrac{4\:+\:8}{2}$

=> $\dfrac{6}{2}$, $\dfrac{8}{2}$, $\dfrac{12}{2}$

=> 3, 4, 6

=> $(\vec{r}\:-\:\vec{a})$ = (x - 3)$\hat{i}$ + (y - 4)$\hat{j}$ + (z - 6)$\hat{k}$

Now..

Normal to plane is

=> $\vec{n}$ = (4 - 2)$\hat{i}$ + (5 - 3)$\hat{j}$ + (8 - 4)$\hat{k}$

=> 2$\hat{i}$ + 2$\hat{j}$ + 4$\hat{k}$

Now..

=> $(\vec{r}\:-\:\vec{a})$.$\vec{n}$ = 0

=> (x - 3)$\hat{i}$ + (y - 4)$\hat{j}$ + (z - 6)$\hat{k}$ . 2$\hat{i}$ + 2$\hat{j}$ + 4$\hat{k}$

=> 2x - 6 + 2y - 8 + 4z - 24 = 0

=> 2x + 2y + 4z - 38 = 0

=> 2(x + y + 2z - 19) = 0

=> x + y + 2z - 19 = 0

=> x + y + 2z = 19

_____________________________

Answer 2

Answer:

The given points are A(2,3,4)andB(4,5,8)

The line segement AB is given by (x2−x1),(y2−y1),(z2−z1)

(ie)(4−2),(5−3),(8−4)

=(2,2,4)=(1,1,2)

Since the plane bisects AB at rightangles, AB−→− is the normal to the plane. which is n→

Therefore n→=i^+j^+2k^

Let C be the midpoint of AB.

(x1+x22,y1+y22,z1+z22)

=C(2+42,3+52,4+82)

=C(62,82,122)=(3,4,6)

Let this be a→=3i^+4j^+6k^

Hence the vector equation of the plane passing through C and ⊥ AB is

(r→−(3i^+4j^+6k^)).(i^+j^+2k^)=0

We know r→=xi^+yj^+zk^

=>[(xi^+yj^+zk^).(3i^+4j^+6k^)].(i^+j^+2k^)=0

On simplifying we get,

(x−3)i^+(y−4)j^+(z−6)k^).(i^+j^+2k^)=0

Applying the product rule we get,

(x−3)+(y−4)+2(z−6)=0

On simplifying we get,

x+y+2z=19

This is the required equation of the plane