Question

If the foot of the perpendicular from the origin to a plane is (a, b, c), the equation of the plane is.
(A) $\dfrac{x}{\text{a}}+\dfrac{\text{y}}{\text{b}}+\dfrac{\text{z}}{\text{c}}=3$
(B) ax + by + cz = 3
(C) ax + by + cz = ${{\text{a}}^{2}}+{{\text{b}}^{2}}+{{\text{c}}^{2}}$
(D) ax + by + cz = a + b + c

Answer 1

Hint: Hint! Let the foot of perpendicular be p(a, b, c). Then directions ratio of OP are (a − 0, b − 0, c − 0) = (a, b c) then, direction ratios of the normal to a, b, c is
a(x − a) + b(y − b) + c(z − c) = 0

Complete step by step solution: Given that,
the foot of the perpendicular from the origin to a plane is (a, b, c).
Now,
According to the question.
⇒ Let the foot of the perpendicular be P(a, b, c).
Then,
⇒ direction ratios of OP are (a − 0, b − 0, c − 0) = (a, b, c).
So,
⇒ the equation of plane passing through P(a, b, c)
then,
the direction ratios of the normal to which are a, b, c is
$\text{d}\cdot {{\text{r}}_{8}}(x-{{x}_{1}})+\text{d}\cdot \text{r}(\text{y}-{{\text{y}}_{1}})+\text{d}\cdot {{\text{r}}_{8}}(\text{z}-{{\text{z}}_{1}})=0$
where,
${{x}_{1}}$ = x-co-ordinates of point P.
${{\text{y}}_{1}}$ = y-co-ordinates of point P.
${{\text{z}}_{1}}$ = z-co-ordinates of point P.
Now,
⇒ Here, \[{{x}_{1}}=\text{a}\], \[{{\text{y}}_{1}}=\text{b}\], \[{{\text{z}}_{1}}=\text{c}\].
So,
⇒ a(x − a) + b(y − b) + c(z − c) = 0
⇒ After multiplying, we get;
⇒ $\text{a}x-{{\text{a}}^{2}}+\text{by}-{{\text{b}}^{2}}+\text{cz}-{{\text{c}}^{2}}=0$
⇒ ax + by + cz = ${{\text{a}}^{2}}+{{\text{b}}^{2}}+{{\text{c}}^{2}}$.

Hence, the correct option is (C).

Note: In this type of question we know about the direction ratios that is If a, b, c are three numbers proportional to the direction cosine l, m, n of a straight line, then a, b, c are called its direction ratios..then, First, find out the direction ratios of line and then find direction ratios of the normal to point (a, b, c)
= $\text{d}\cdot {{\text{r}}_{8}}(x-{{x}_{1}})+\text{d}\cdot {{\text{r}}_{8}}(\text{y}-{{\text{y}}_{1}})+\text{d}\cdot {{\text{r}}_{8}}(\text{z}-{{\text{z}}_{1}})=0$