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JEE Main · MCQ · 3D Geometry · Lines in Space
MCQ · Mathematics · 3D Geometry
Q. Let the line $L$ pass through the point $(-3,\,5,\,2)$ and make equal angles with the positive coordinate axes. If the distance of $L$ from the point $(-2,\,r,\,1)$ is $\sqrt{\dfrac{14}{3}}$, then the sum of all possible values of $r$ is:
A$16$
B$12$
C$6$
D$10$ ✓
✅ Correct Answer: (D) $10$
Step-by-Step Solution
1
Direction cosines of line $L$ making equal angles with positive axes
If $L$ makes equal angles $\alpha$ with the positive $x$-, $y$-, $z$-axes, then $l = m = n = \cos\alpha$.
Using the identity $l^2 + m^2 + n^2 = 1$:
Using the identity $l^2 + m^2 + n^2 = 1$:
$$\cos^2\alpha + \cos^2\alpha + \cos^2\alpha = 1$$
$$3\cos^2\alpha = 1 \implies \cos\alpha = \frac{1}{\sqrt{3}}$$
Direction ratios of $L$: $\;(1,\,1,\,1)$
Unit direction vector: $\;\hat{d} = \dfrac{1}{\sqrt{3}}(1,\,1,\,1)$
2
Set up the distance formula from point to line in 3D
Let $A = (-3,\,5,\,2)$ be the point on $L$, and $P = (-2,\,r,\,1)$ be the external point.
$$\vec{AP} = (-2-(-3),\; r-5,\; 1-2) = (1,\; r-5,\; -1)$$
Distance formula:
$$d^2 = |\vec{AP}|^2 – \left(\vec{AP}\cdot\hat{d}\right)^2$$
3
Compute $|\vec{AP}|^2$
$$|\vec{AP}|^2 = 1^2 + (r-5)^2 + (-1)^2 = 2 + (r-5)^2$$
4
Compute $\vec{AP} \cdot \hat{d}$
$$\vec{AP}\cdot\hat{d} = \frac{1}{\sqrt{3}}\bigl(1\cdot1 + (r-5)\cdot1 + (-1)\cdot1\bigr)$$
$$= \frac{1}{\sqrt{3}}\bigl(1 + r – 5 – 1\bigr) = \frac{r-5}{\sqrt{3}}$$
Therefore:
$$\left(\vec{AP}\cdot\hat{d}\right)^2 = \frac{(r-5)^2}{3}$$
5
Apply $d^2 = \dfrac{14}{3}$ and form the equation
$$2 + (r-5)^2 – \frac{(r-5)^2}{3} = \frac{14}{3}$$
Let $u = (r-5)^2$. Simplify:
$$2 + u – \frac{u}{3} = \frac{14}{3}$$
$$2 + \frac{2u}{3} = \frac{14}{3}$$
$$\frac{2u}{3} = \frac{14}{3} – 2 = \frac{14-6}{3} = \frac{8}{3}$$
$$u = \frac{8}{3} \times \frac{3}{2} = 4$$
6
Solve $(r-5)^2 = 4$
$$r – 5 = \pm 2$$
$$r = 5 + 2 = 7 \quad \text{or} \quad r = 5 – 2 = 3$$
Two possible values: $r = 7$ and $r = 3$
7
Sum of all possible values of $r$
$$r_1 + r_2 = 7 + 3 = 10$$
Sum $= 10$ → Option (D) ✓
Related Theory
📌 Direction Cosines and Direction Ratios — Complete Theory
If a line makes angles $\alpha,\,\beta,\,\gamma$ with the positive $x$-, $y$-, $z$-axes respectively, then:
Direction cosines (DCs): $l = \cos\alpha,\; m = \cos\beta,\; n = \cos\gamma$
Fundamental identity: $$l^2 + m^2 + n^2 = 1$$ This is always true for direction cosines — it comes from the fact that the direction vector has unit length.
Direction ratios (DRs): Any three numbers $a,b,c$ proportional to $l,m,n$ are DRs. If DRs are $(a,b,c)$: $$l = \frac{a}{\sqrt{a^2+b^2+c^2}}, \quad m = \frac{b}{\sqrt{a^2+b^2+c^2}}, \quad n = \frac{c}{\sqrt{a^2+b^2+c^2}}$$
Direction cosines (DCs): $l = \cos\alpha,\; m = \cos\beta,\; n = \cos\gamma$
Fundamental identity: $$l^2 + m^2 + n^2 = 1$$ This is always true for direction cosines — it comes from the fact that the direction vector has unit length.
Direction ratios (DRs): Any three numbers $a,b,c$ proportional to $l,m,n$ are DRs. If DRs are $(a,b,c)$: $$l = \frac{a}{\sqrt{a^2+b^2+c^2}}, \quad m = \frac{b}{\sqrt{a^2+b^2+c^2}}, \quad n = \frac{c}{\sqrt{a^2+b^2+c^2}}$$
$l^2+m^2+n^2=1$
DRs $(a,b,c)$: $l=a/\sqrt{a^2+b^2+c^2}$
📌 Lines Making Equal Angles with Axes — Four Possibilities
“Equal angles with positive coordinate axes” strictly means $l = m = n > 0$, giving only one direction: $(1,1,1)/\sqrt{3}$.
However, “equal angles with coordinate axes” (without “positive”) allows four possibilities since each cosine can be $\pm 1/\sqrt{3}$:
In this problem, since “positive axes” is specified, only $(1,1,1)$ is used. The distance from a point to a line is the same regardless of which direction along the line is chosen, so $(1,1,1)$ and $(-1,-1,-1)$ give the same distance.
However, “equal angles with coordinate axes” (without “positive”) allows four possibilities since each cosine can be $\pm 1/\sqrt{3}$:
| Direction Ratios | $l$ | $m$ | $n$ |
|---|---|---|---|
| $(1,1,1)$ | $+1/\sqrt{3}$ | $+1/\sqrt{3}$ | $+1/\sqrt{3}$ |
| $(1,1,-1)$ | $+1/\sqrt{3}$ | $+1/\sqrt{3}$ | $-1/\sqrt{3}$ |
| $(1,-1,1)$ | $+1/\sqrt{3}$ | $-1/\sqrt{3}$ | $+1/\sqrt{3}$ |
| $(-1,1,1)$ | $-1/\sqrt{3}$ | $+1/\sqrt{3}$ | $+1/\sqrt{3}$ |
📌 Distance from a Point to a Line in 3D
Given a line passing through point $A$ with unit direction vector $\hat{d}$, the distance from external point $P$ to the line is:
$$d = \sqrt{|\vec{AP}|^2 – (\vec{AP}\cdot\hat{d})^2}$$
Or equivalently using cross product:
$$d = |\vec{AP} \times \hat{d}|$$
Derivation: Project $\vec{AP}$ onto the line direction $\hat{d}$. The projection length is $\vec{AP}\cdot\hat{d}$. The perpendicular distance is the remaining component by Pythagoras.
Why these are equivalent: $|\vec{AP}\times\hat{d}|^2 = |\vec{AP}|^2|\hat{d}|^2 – (\vec{AP}\cdot\hat{d})^2 = |\vec{AP}|^2 – (\vec{AP}\cdot\hat{d})^2$ since $|\hat{d}|=1$.
Derivation: Project $\vec{AP}$ onto the line direction $\hat{d}$. The projection length is $\vec{AP}\cdot\hat{d}$. The perpendicular distance is the remaining component by Pythagoras.
Why these are equivalent: $|\vec{AP}\times\hat{d}|^2 = |\vec{AP}|^2|\hat{d}|^2 – (\vec{AP}\cdot\hat{d})^2 = |\vec{AP}|^2 – (\vec{AP}\cdot\hat{d})^2$ since $|\hat{d}|=1$.
$d = \sqrt{|\vec{AP}|^2 – (\vec{AP}\cdot\hat{d})^2}$
$d = |\vec{AP}\times\hat{d}|$
📌 Equation of a Line in 3D — Symmetric Form
A line passing through $(x_1,y_1,z_1)$ with direction ratios $(a,b,c)$:
$$\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} = \lambda$$
In this problem, line $L$:
$$\frac{x+3}{1} = \frac{y-5}{1} = \frac{z-2}{1} = \lambda$$
Any point on $L$: $(\lambda-3,\; \lambda+5,\; \lambda+2)$
The foot of perpendicular from $P(-2,r,1)$ to $L$ is found by: $$\vec{AP_0}\cdot\hat{d} = 0 \quad \text{where } P_0 \text{ is foot}$$ But using the distance formula directly is faster for this problem.
The foot of perpendicular from $P(-2,r,1)$ to $L$ is found by: $$\vec{AP_0}\cdot\hat{d} = 0 \quad \text{where } P_0 \text{ is foot}$$ But using the distance formula directly is faster for this problem.
Symmetric form: $(x-x_1)/a=(y-y_1)/b=(z-z_1)/c$
Point on line: $(x_1+a\lambda, y_1+b\lambda, z_1+c\lambda)$
📌 Foot of Perpendicular Method (Alternative Approach)
An alternative to the distance formula: find the foot of perpendicular $F$ from $P(-2,r,1)$ to line $L$.
Any point on $L$: $F = (\lambda-3,\; \lambda+5,\; \lambda+2)$
$\vec{FP} = (-2-(\lambda-3),\; r-(\lambda+5),\; 1-(\lambda+2)) = (1-\lambda,\; r-5-\lambda,\; -1-\lambda)$
For perpendicularity: $\vec{FP}\cdot(1,1,1) = 0$: $$(1-\lambda) + (r-5-\lambda) + (-1-\lambda) = 0$$ $$r – 5 – 3\lambda = 0 \implies \lambda = \frac{r-5}{3}$$ Then $|FP|^2 = \dfrac{14}{3}$ gives the same quadratic in $r$.
Any point on $L$: $F = (\lambda-3,\; \lambda+5,\; \lambda+2)$
$\vec{FP} = (-2-(\lambda-3),\; r-(\lambda+5),\; 1-(\lambda+2)) = (1-\lambda,\; r-5-\lambda,\; -1-\lambda)$
For perpendicularity: $\vec{FP}\cdot(1,1,1) = 0$: $$(1-\lambda) + (r-5-\lambda) + (-1-\lambda) = 0$$ $$r – 5 – 3\lambda = 0 \implies \lambda = \frac{r-5}{3}$$ Then $|FP|^2 = \dfrac{14}{3}$ gives the same quadratic in $r$.
📌 Cross Product Method for Distance
$d = |\vec{AP}\times\hat{d}|$ is often faster when $\hat{d}$ has a simple form.
Here $\vec{AP} = (1,\,r-5,\,-1)$ and $\hat{d} = \dfrac{1}{\sqrt{3}}(1,1,1)$: $$\vec{AP}\times(1,1,1) = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&r-5&-1\\1&1&1\end{vmatrix}$$ $$= \mathbf{i}[(r-5)(1)-(-1)(1)] – \mathbf{j}[(1)(1)-(-1)(1)] + \mathbf{k}[(1)(1)-(r-5)(1)]$$ $$= \mathbf{i}(r-4) – \mathbf{j}(2) + \mathbf{k}(6-r)$$ $$= (r-4,\,-2,\,6-r)$$ $$|\vec{AP}\times(1,1,1)|^2 = (r-4)^2 + 4 + (6-r)^2$$ Since $d = |\vec{AP}\times\hat{d}| = |\vec{AP}\times(1,1,1)|/\sqrt{3}$: $$d^2 = \frac{(r-4)^2+4+(6-r)^2}{3} = \frac{14}{3}$$ $$(r-4)^2+(6-r)^2 = 10$$ Expanding: $r^2-8r+16+r^2-12r+36=10 \implies 2r^2-20r+42=0 \implies r^2-10r+21=0$ $$(r-7)(r-3)=0 \implies r=7 \text{ or } r=3$$ Sum $= 10$ ✓
Here $\vec{AP} = (1,\,r-5,\,-1)$ and $\hat{d} = \dfrac{1}{\sqrt{3}}(1,1,1)$: $$\vec{AP}\times(1,1,1) = \begin{vmatrix}\mathbf{i}&\mathbf{j}&\mathbf{k}\\1&r-5&-1\\1&1&1\end{vmatrix}$$ $$= \mathbf{i}[(r-5)(1)-(-1)(1)] – \mathbf{j}[(1)(1)-(-1)(1)] + \mathbf{k}[(1)(1)-(r-5)(1)]$$ $$= \mathbf{i}(r-4) – \mathbf{j}(2) + \mathbf{k}(6-r)$$ $$= (r-4,\,-2,\,6-r)$$ $$|\vec{AP}\times(1,1,1)|^2 = (r-4)^2 + 4 + (6-r)^2$$ Since $d = |\vec{AP}\times\hat{d}| = |\vec{AP}\times(1,1,1)|/\sqrt{3}$: $$d^2 = \frac{(r-4)^2+4+(6-r)^2}{3} = \frac{14}{3}$$ $$(r-4)^2+(6-r)^2 = 10$$ Expanding: $r^2-8r+16+r^2-12r+36=10 \implies 2r^2-20r+42=0 \implies r^2-10r+21=0$ $$(r-7)(r-3)=0 \implies r=7 \text{ or } r=3$$ Sum $= 10$ ✓
Cross product confirms: $r=3$ or $r=7$
Sum $= 10$ ✓
📌 Common Mistakes to Avoid
❌ Mistake 1: Using direction ratios $(1,1,1)$ without converting to unit vector $\hat{d} = (1,1,1)/\sqrt{3}$ when computing $\vec{AP}\cdot\hat{d}$.
❌ Mistake 2: Computing $|\vec{AP}|^2 – (\vec{AP}\cdot\vec{d})^2$ using non-unit $\vec{d}$. Always use unit direction vector.
❌ Mistake 3: Setting $d = \sqrt{14/3}$ and computing $d^2 = 14/\sqrt{3}$ — remember $(\sqrt{14/3})^2 = 14/3$.
❌ Mistake 4: Finding only one value of $r$. The equation $(r-5)^2=4$ gives two values — always check for both roots.
❌ Mistake 5: Using the 2D distance formula instead of the 3D formula.
❌ Mistake 2: Computing $|\vec{AP}|^2 – (\vec{AP}\cdot\vec{d})^2$ using non-unit $\vec{d}$. Always use unit direction vector.
❌ Mistake 3: Setting $d = \sqrt{14/3}$ and computing $d^2 = 14/\sqrt{3}$ — remember $(\sqrt{14/3})^2 = 14/3$.
❌ Mistake 4: Finding only one value of $r$. The equation $(r-5)^2=4$ gives two values — always check for both roots.
❌ Mistake 5: Using the 2D distance formula instead of the 3D formula.
📌 Key Formulas Summary
$l^2+m^2+n^2=1$
Equal angles with axes: $l=m=n=1/\sqrt{3}$
$d^2=|\vec{AP}|^2-(\vec{AP}\cdot\hat{d})^2$
$d=|\vec{AP}\times\hat{d}|$
$(r-5)^2=4 \Rightarrow r=3$ or $7$
Sum of roots $= 10$
Frequently Asked Questions
1. What are direction cosines of a line making equal angles with positive axes?
$l=m=n=1/\sqrt{3}$, so direction ratios are $(1,1,1)$.
2. What is $\hat{d}$?
$\hat{d} = (1,1,1)/\sqrt{3}$.
3. What is $\vec{AP}$?
$(-2+3,\;r-5,\;1-2) = (1,\;r-5,\;-1)$.
4. What is $|\vec{AP}|^2$?
$1+(r-5)^2+1 = 2+(r-5)^2$.
5. What is $\vec{AP}\cdot\hat{d}$?
$(1+r-5-1)/\sqrt{3} = (r-5)/\sqrt{3}$.
6. What equation is formed?
$(r-5)^2 = 4$.
7. What are the two values of $r$?
$r=7$ and $r=3$.
8. What is the sum of values of $r$?
$7+3=10$.
9. What identity do direction cosines satisfy?
$l^2+m^2+n^2=1$ always.
10. Which distance formula is used?
$d^2 = |\vec{AP}|^2 – (\vec{AP}\cdot\hat{d})^2$.
11. Can cross product method also be used?
Yes. $d = |\vec{AP}\times\hat{d}|$ gives the same result: $r=3$ or $7$, sum $=10$.
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