Let A, B, C be the feet of perpendiculars from a point P on the $x$, $y$, $z$-axis respectively. Find the coordinates of A, B and C in each of the following where the point P is: (i) $A=(3,4,2)$, (ii) $(-5,3,7)$, (iii) $(4,-3,-5)$.

Let A, B, C be the feet of perpendiculars from a point P on the x, y, z-axis respectively. Find the coordinate

NCERT Exemplar · Q3

x

(i)

A=(3,4,2)

(ii)

(-5,3,7)

(iii)

(4,-3,-5)

Chhattisgarh CgbseShort· 2mImportance★★★★★

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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}.

Its magnitude is $\sqrt{25+36+4}=\sqrt{65}$ , and $|\vec{b}|=\sqrt{4+1+4}=3$ , so

d = \frac{\sqrt{65}}{3}.

Watch out

$\vec{b}$ must be the line's direction vector, not a point on the line. And use $\overrightarrow{AP}=\vec{p}-\vec{a}$ where $A$ is any point genuinely on the line.

Finding the shortest distance from a point to a line using the cross product is a standard, frequently tested problem in the NCERT Class 12 Three Dimensional Geometry chapter, appearing in CBSE boards, JEE Main and various state CETs. "Distance of a point from a line vector form" is a common search, and this same cross-product technique reappears later when finding the distance between two skew lines.

The foot of perpendicular from a point to a coordinate axis is found by dropping the other two coordinates to zero; for the $x$ -axis keep only the $x$ -coordinate, for the $y$ -axis keep only $y$ , and for the $z$ -axis keep only $z$ . The feet are $A = (x, 0, 0)$ , $B = (0, y, 0)$ , $C = (0, 0, z)$ for point $P = (x, y, z)$ .

Understanding perpendiculars to coordinate axes

When we drop a perpendicular from a point $P$ in three-dimensional space to one of the coordinate axes, we're finding the closest point on that axis to $P$ . The key geometric insight is that each coordinate axis is characterized by two coordinates being zero.

The $x$ -axis consists of all points of the form $(x, 0, 0)$ . The $y$ -axis has points $(0, y, 0)$ . The $z$ -axis contains points $(0, 0, z)$ .

When you drop a perpendicular from $P = (x_0, y_0, z_0)$ to, say, the $x$ -axis, you're moving in a direction perpendicular to that axis. The $x$ -axis points along the vector $(1, 0, 0)$ , so any perpendicular motion must be in the $yz$ -plane. This means the $x$ -coordinate stays fixed while $y$ and $z$ change. The foot of the perpendicular is therefore $(x_0, 0, 0)$ .

Tip

To find the foot of perpendicular to a coordinate axis, simply "zero out" the coordinates corresponding to the other two axes.

Solution for each case

1. For $P = (3, 4, 2)$ :

The foot on the $x$ -axis keeps the $x$ -coordinate and zeros the rest:

A = (3, 0, 0)

The foot on the $y$ -axis keeps the $y$ -coordinate:

B = (0, 4, 0)

The foot on the $z$ -axis keeps the $z$ -coordinate:

C = (0, 0, 2)

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

Applying the same principle:

A = (-5, 0, 0)

B = (0, 3, 0)

C = (0, 0, 7)

3. For $P = (4, -3, -5)$ :

Again, preserving one coordinate at a time:

A = (4, 0, 0)

B = (0, -3, 0)

C = (0, 0, -5)

Note

Notice that negative coordinates are preserved as-is. The foot of perpendicular from $(4, -3, -5)$ to the $y$ -axis is $(0, -3, 0)$ , not $(0, 3, 0)$ .

✓Final answer

P = (3, 4, 2)

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Concept understanding — Distance From Point To Line

Distance from a Point to a Line

The idea

Example

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Understanding perpendiculars to coordinate axes

Solution for each case

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Solution for each case

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Q.Let A, B, C be the feet of perpendiculars from a point P on the $x$ , $y$ , $z$ -axis respectively. Find the coordinates of A, B and C in each of the following where the point P is:

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