Question

If from a point \[P\left( {a,b,c} \right)\] perpendiculars \[PA\] and \[PB\] are drawn to \[yz\] and \[zx\] planes. Then find the equation of the plane \[OAB\], where \[O\] is the origin.
A. \[bcx + cay + abz = 0\]
B. \[bcx + cay - abz = 0\]
C. \[bcx - cay + abz = 0\]
D. \[ - bcx + cay + abz = 0\]

Answer 1

Hint: Here, the co-ordinates of the point are given. First, calculate the co-ordinates of the points \[A\] and \[B\] on the basis of the given information. Then, calculate the equation of the plane passing through the origin and a point. After that, substitute the co-ordinates of the points \[A\] and \[B\] in the equation of the plane and solve them. Then, substitute the values of the direction ratios in the equation of the plane and solve it to get the required answer.

Formula used: The equation of a plane passing through the point \[\left( {{x_1},{y_1},{z_1}} \right)\] with direction ratios \[\left( {a,b,c} \right)\]: \[a\left( {x - {x_1}} \right) + b\left( {y - {y_1}} \right) + c\left( {z - {z_1}} \right) = 0\]

Complete step by step solution: Given:
\[PA\] and \[PB\] are the perpendiculars drawn from the point \[P\left( {a,b,c} \right)\] to \[yz\] and \[zx\] planes.
The plane \[OAB\] passes through the origin.

Since \[PA\] is perpendicular to the \[yz\] plane.
So, \[x\] co-ordinate of the point \[A\] is 0.
Thus, the co-ordinates of the point \[A\] are: \[\left( {0,b,c} \right)\]

Similarly, \[PB\] is perpendicular to the \[zx\] plane.
So, \[y\] co-ordinate of the point \[B\] is 0.
Thus, the co-ordinates of the point \[B\] are: \[\left( {a,0,c} \right)\]

It is given that the plane \[OAB\] passes through the origin.
Let consider, \[p,q,r\] are the direction ratios of the plane.
Then, the equation of the plane is:
\[p\left( {x - 0} \right) + q\left( {y - 0} \right) + r\left( {z - 0} \right) = 0\]
\[ \Rightarrow px + qy + rz = 0\] \[.....\left( 1 \right)\]

Also. The equation passes through the points \[A\left( {0,b,c} \right)\] and \[B\left( {a,0,c} \right)\].
So, we get
\[p\left( 0 \right) + q\left( b \right) + r\left( c \right) = 0\]
\[ \Rightarrow qb + rc = 0\] \[.....\left( 2 \right)\]

\[p\left( a \right) + q\left( 0 \right) + r\left( c \right) = 0\]
\[ \Rightarrow pa + rc = 0\] \[.....\left( 3 \right)\]

Solving equations \[\left( 2 \right)\] and \[\left( 3 \right)\], we get
\[\dfrac{p}{{bc}} = \dfrac{q}{{ac}} = \dfrac{r}{{ - ab}} = k\]
\[ \Rightarrow p = kbc,q = kac,r = - kab\]
Now substitute the above values in the equation \[\left( 1 \right)\].
We get,
\[kbcx + kacy - kabz = 0\]
\[ \Rightarrow k\left( {bcx + acy - abz} \right) = 0\]
\[ \Rightarrow bcx + acy - abz = 0\]
Thus, the equation of the plane \[OAB\] is \[bcx + acy - abz = 0\].

Thus, Option (B) is correct.

Note: Remember the following forms of the equation of a plane:
General equation of a plane: \[ax + by + cz + d = 0\] , where \[d \ne 0\]
Equation of plane whose intercepts are \[a,b,c\] is: \[\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1\]
Equation of a plane passing through the point \[\left( {{x_1},{y_1},{z_1}} \right)\] with direction ratios \[\left( {a,b,c} \right)\]: \[a\left( {x - {x_1}} \right) + b\left( {y - {y_1}} \right) + c\left( {z - {z_1}} \right) = 0\]