If the distance between the points (a,2,1) and (1,-1,1) is 5, then a ____.

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(a,2,1)

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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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Use the three-dimensional distance formula between two points; equating the distance to 5 gives a quadratic equation in $a$ with two solutions: $a = 5$ or $a = -3$ .

The distance between two points in three-dimensional space follows directly from the Pythagorean theorem extended to three dimensions. If you imagine the two points as opposite corners of a rectangular box aligned with the coordinate axes, the distance is the length of the diagonal through that box.

For points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ , the distance formula is:

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

This measures how far apart the points are in each coordinate direction, then combines those separations into a single straight-line distance.

Let me apply this to find $a$ .

Identify the coordinates and given distance.

The first point is $(a, 2, 1)$ and the second is $(1, -1, 1)$ . We know the distance between them is $5$ .
Compute the differences in each coordinate.
- In the $x$ -direction: $1 - a$
- In the $y$ -direction: $-1 - 2 = -3$
- In the $z$ -direction: $1 - 1 = 0$
Apply the distance formula.

\sqrt{(1-a)^2 + (-3)^2 + 0^2} = 5

Simplifying:

\sqrt{(1-a)^2 + 9} = 5

Square both sides to eliminate the square root.

(1-a)^2 + 9 = 25

$(1-a)^2 = 16$ …

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Q.If the distance between the points (a,2,1)(a,2,1)(a,2,1) and (1,−1,1)(1,-1,1)(1,−1,1) is 5, then aaa _______.

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Q.If the distance between the points $(a,2,1)$ and $(1,-1,1)$ is 5, then $a$ _______.

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