Distance of the point $(3,4,5)$ from the origin $(0,0,0)$ is (A) $\sqrt{50}$ (B) $3$ (C) $4$ (D) $5$

Distance of the point (3,4,5) from the origin (0,0,0) is (A) √(50) (B) 3 (C) 4 (D) 5

NCERT Exemplar · Q24

(3,4,5)

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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 distance between two points in three-dimensional space is found using the natural extension of the Pythagorean theorem; here the distance from $(3,4,5)$ to the origin is $\sqrt{3^2+4^2+5^2} = \sqrt{50} = 5\sqrt{2}$ .

Why the distance formula works

When we move from the plane to three-dimensional space, the idea of distance remains rooted in the Pythagorean theorem. Imagine standing at the origin and wanting to reach the point $(3,4,5)$ . You could walk 3 units along the $x$ -axis, then 4 units parallel to the $y$ -axis, and finally 5 units parallel to the $z$ -axis. These three perpendicular displacements form the edges of a rectangular box, and the straight-line distance is the space diagonal of that box.

The Pythagorean theorem applied twice gives us the formula: first in the $xy$ -plane to get $\sqrt{3^2+4^2}$ , then combining that with the $z$ -component to get $\sqrt{(\sqrt{3^2+4^2})^2 + 5^2} = \sqrt{3^2+4^2+5^2}$ .

d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}

Step-by-step calculation

Identify the coordinates. We have point $P = (3,4,5)$ and the origin $O = (0,0,0)$ .
Apply the distance formula. The distance is

d = \sqrt{(3-0)^2 + (4-0)^2 + (5-0)^2}

Compute each squared term:
- $(3-0)^2 = 9$
- $(4-0)^2 = 16$ …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Higher Secondary (+2 Stage) Examination 2025Set ANNUAL5 marks
$(1,0,0)$
›Reveal solutionSolution

Write a general point on the line, force the vector from the given point to it to be perpendicular to the line's direction, solve for the parameter, then read off the foot, distance, and perpendicular's equation.

The line is $\dfrac{x-1}{2}=\dfrac{y+1}{-3}=\dfrac{z+10}{8}=t$ , passing through $(1,-1,-10)$ with direction ratios $(2,-3,8)$ . A general point on it is

$Q=(1+2t,\,-1-3t,\,-10+8t).$

The vector from the given point $P(1,0,0)$ to $Q$ is

$\vec{PQ}=(2t,\;-1-3t,\;-10+8t).$

For $Q$ to be the foot of the perpendicular, $\vec{PQ}$ must be perpendicular to the line's direction $(2,-3,8)$ , i.e. their dot product is zero:

$2(2t)+(-3)(-1-3t)+8(-10+8t)=0$

$4t+3+9t-80+64t=0$

$77t-77=0\ \Rightarrow\ t=1.$

Foot of perpendicular: substituting $t=1$ : $Q=(1+2,\,-1-3,\,-10+8)=(3,-4,-2)$ .

Distance: $\vec{PQ}=(3-1,\,-4-0,\,-2-0)=(2,-4,-2)$ , so

$|PQ|=\sqrt{2^2+(-4)^2+(-2)^2}=\sqrt{4+16+4}=\sqrt{24}=2\sqrt6.$

…

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Higher Secondary (+2 Stage) Examination 2023Set ANNUAL5 marks
$(1,6,3)$
›Reveal solutionSolution

Find the foot of the perpendicular from the point to the line (by making the vector from a general point on the line to the given point perpendicular to the line's direction), then the image is the point such that this foot is the midpoint of the given point and its image.

Line: $\dfrac{x}{1}=\dfrac{y-1}{2}=\dfrac{z-2}{3}=t$ , so a general point on the line is $Q(t)=(t,\,1+2t,\,2+3t)$ , with direction ratios $(1,2,3)$ .

Let $P=(1,6,3)$ .

Find the foot of the perpendicular $F$ : The vector $\overrightarrow{QP} = (1-t,\ 6-(1+2t),\ 3-(2+3t)) = (1-t,\ 5-2t,\ 1-3t)$ must be perpendicular to the direction $(1,2,3)$ :

$(1-t)(1)+(5-2t)(2)+(1-3t)(3)=0$

$1-t+10-4t+3-9t=0$

$14-14t=0 \;\Rightarrow\; t=1$

So the foot of the perpendicular is $F=Q(1)=(1,\,1+2,\,2+3)=(1,3,5)$ .

Find the image $P'$ : Since $F$ is the midpoint of $P$ and its image $P'$ :

$F=\dfrac{P+P'}{2} \;\Rightarrow\; P'=2F-P$

$P' = (2(1)-1,\ 2(3)-6,\ 2(5)-3) = (1,\,0,\,7)$

…

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Q.Distance of the point (3,4,5)(3,4,5)(3,4,5) from the origin (0,0,0)(0,0,0)(0,0,0) is (A) 50\sqrt{50}50​ (B) 333 (C) 444 (D) 555

Why the distance formula works

Step-by-step calculation

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Q.Distance of the point $(3,4,5)$ from the origin $(0,0,0)$ is
(A) $\sqrt{50}$
(B) $3$
(C) $4$
(D) $5$

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