Let A, B, C be the feet of perpendiculars from a point P on the $xy$, $yz$ and $zx$-planes respectively. Find the coordinates of A, B, C in each of the following where the point P is: (i) $(3,4,5)$, (ii) $(-5,3,7)$, (iii) $(4,-3,-5)$.

Let A, B, C be the feet of perpendiculars from a point P on the xy, yz and zx-planes respectively. Find the co

NCERT Exemplar · Q4

xy

(i)

(3,4,5)

(ii)

(-5,3,7)

(iii)

(4,-3,-5)

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Concept understanding — Distance From Point To Line

Distance from a Point to a Line

The distance from a point to a line is the shortest distance — the length of the perpendicular dropped from the point onto the line. In 3D we compute it with vectors and the cross product.

Let the line be $\vec{r}=\vec{a}+\lambda\vec{b}$ (a point $A$ with position vector $\vec{a}$ , direction $\vec{b}$ ), and let $P$ be the given point with position vector $\vec{p}$ .

The idea

Look at the triangle formed by $A$ , $P$ and the foot of the perpendicular $M$ . The segment $\overrightarrow{AP}=\vec{p}-\vec{a}$ is the hypotenuse, and the perpendicular distance $d=PM$ is the side opposite the angle $\theta$ between $\overrightarrow{AP}$ and the line:

d = |\overrightarrow{AP}|\sin\theta.

But the cross product already contains $\sin\theta$ : $|\overrightarrow{AP}\times\vec{b}| = |\overrightarrow{AP}|\,|\vec{b}|\sin\theta$ . Dividing by $|\vec{b}|$ isolates the distance.

d = \frac{|(\vec{p}-\vec{a})\times\vec{b}|}{|\vec{b}|}

Example

Distance of $P(1,2,3)$ from the line $\vec{r}=(\hat{i}+\hat{j})+\lambda(2\hat{i}-\hat{j}+2\hat{k})$ .

Here $\vec{a}=(1,1,0)$ , $\vec{b}=(2,-1,2)$ , and $\overrightarrow{AP}=\vec{p}-\vec{a}=(0,1,3)$ .

$\overrightarrow{AP}\times\vec{b} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\0&1&3\\2&-1&2\end{vmatrix} = (2+3)\hat{i} - (0-6)\hat{j} + (0-2)\hat{k} = 5\hat{i}+6\hat{j}-2\hat{k}.$ …

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The foot of the perpendicular from a point to a coordinate plane is found by setting the coordinate perpendicular to that plane to zero, while keeping the other two coordinates unchanged. For P(3,4,5), the feet are A(3,4,0), B(0,4,5), C(3,0,5); for P(-5,3,7), A(-5,3,0), B(0,3,7), C(-5,0,7); for P(4,-3,-5), A(4,-3,0), B(0,-3,-5), C(4,0,-5).

The core idea: dropping a perpendicular onto a coordinate plane

When you drop a perpendicular from a point to a plane, the foot is the point on the plane that lies directly "below" or "above" the original point along the direction perpendicular to that plane.

For a coordinate plane, this is beautifully simple. The $xy$ -plane is the set of all points where $z = 0$ . The direction perpendicular to it is the $z$ -axis. So to go from any point straight down to the $xy$ -plane, you keep $x$ and $y$ exactly as they are, and set $z = 0$ .

Similarly, the $yz$ -plane has equation $x = 0$ , so you keep $y$ and $z$ and set $x = 0$ . The $zx$ -plane has equation $y = 0$ , so you keep $x$ and $z$ and set $y = 0$ .

P(x_0, y_0, z_0)

That's the entire logic. No distance formula needed — it's purely geometric reasoning about what "perpendicular to a coordinate plane" means.

Applying it to each point

Let's work through each case.

1. Point P = $(3, 4, 5)$

Foot A on $xy$ -plane: Keep $x=3$ , $y=4$ , set $z=0$ . So $A = (3, 4, 0)$ .
Foot B on $yz$ -plane: Keep $y=4$ , $z=5$ , set $x=0$ . So $B = (0, 4, 5)$ .
Foot C on $zx$ -plane: Keep $x=3$ , $z=5$ , set $y=0$ . So $C = (3, 0, 5)$ .

2. Point P = $(-5, 3, 7)$

Foot A on $xy$ -plane: $A = (-5, 3, 0)$ .
Foot B on $yz$ -plane: $B = (0, 3, 7)$ . …

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Q.Let A, B, C be the feet of perpendiculars from a point P on the xyxyxy, yzyzyz and zxzxzx-planes respectively. Find the coordinates of A, B, C in each of the following where the point P is:

The core idea: dropping a perpendicular onto a coordinate plane

Applying it to each point

1. Point P = (3,4,5)(3, 4, 5)(3,4,5)

2. Point P = (−5,3,7)(-5, 3, 7)(−5,3,7)

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Q.Let A, B, C be the feet of perpendiculars from a point P on the $xy$ , $yz$ and $zx$ -planes respectively. Find the coordinates of A, B, C in each of the following where the point P is:

1. Point P = $(3, 4, 5)$

2. Point P = $(-5, 3, 7)$

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