3D Geometry

3D Geometry

🟢 Lite — Quick Review (1h–1d)

Rapid summary for last-minute revision before your exam.

• Direction cosines of a line through $(x_1, y_1, z_1)$ with direction ratios $a, b, c$ are $l = a/\sqrt{a^2+b^2+c^2}$, $m = b/\sqrt{a^2+b^2+c^2}$, $n = c/\sqrt{a^2+b^2+c^2}$, satisfying the identity $l^2 + m^2 + n^2 = 1$.
• Symmetric form of a line: $\dfrac{x-x_1}{a} = \dfrac{y-y_1}{b} = \dfrac{z-z_1}{c}$, derived from the vector form $\vec{r} = \vec{a} + \lambda \vec{b}$.
• Angle between two lines with direction cosines $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$: $\cos\theta = l_1 l_2 + m_1 m_2 + n_1 n_2$.
• Plane in intercept form: $\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1$, where $a, b, c$ are intercepts on the axes.
• Point-to-plane distance: $d = \dfrac{|Ax_0+By_0+Cz_0+D|}{\sqrt{A^2+B^2+C^2}}$.
• Skew-line distance: $d = \dfrac{|(\vec{b_1}\times\vec{b_2})\cdot(\vec{a_2}-\vec{a_1})|}{|\vec{b_1}\times\vec{b_2}|}$.
• CUET traps: use direction cosines (not ratios) for the angle formula, and never confuse parallel lines (denominator zero) with intersecting lines.

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🟡 Standard — Regular Study (2d–2mo)

Standard content for students with a few days to months.

Direction Cosines and Direction Ratios

For a directed line in space, the direction ratios $(a, b, c)$ are any three numbers proportional to the components of the line’s direction vector. The direction cosines $(l, m, n)$ are these ratios normalised so that they represent cosines of the angles the line makes with the positive x, y and z axes respectively. Hence $l^2 + m^2 + n^2 = 1$ always, while $a, b, c$ need not satisfy this. If $(a, b, c)$ are direction ratios, then $(\pm a, \pm b, \pm c)$ are also direction ratios, but the corresponding direction cosines change sign consistently.

Equation of a Straight Line

A line through point $A(x_1, y_1, z_1)$ with direction ratios $a, b, c$ is written in symmetric (Cartesian) form as

$$\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} = \lambda.$$

The same line in vector form is $\vec{r} = \vec{a} + \lambda \vec{b}$, where $\vec{a} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}$ and $\vec{b} = a\hat{i} + b\hat{j} + c\hat{k}$. Two lines are parallel when their direction ratios are proportional and perpendicular when $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$.

Equation of a Plane

Form	Equation	Parameters
Normal	$lx + my + nz = d$	$(l, m, n)$ = unit normal
Cartesian	$Ax + By + Cz + D = 0$	Normal = $(A, B, C)$
Intercept	$\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1$	Intercepts on axes
Through three points	$\begin{vmatrix} x-x_1 & y-y_1 & z-z_1 \ x_2-x_1 & y_2-y_1 & z_2-z_1 \ x_3-x_1 & y_3-y_1 & z_3-z_1 \end{vmatrix} = 0$	Uses coplanarity

Angles in 3D

• Between two lines: $\cos\theta = l_1l_2 + m_1m_2 + n_1n_2$.
• Between a line and a plane: $\sin\theta = \dfrac{|aA+bB+cC|}{\sqrt{a^2+b^2+c^2}\sqrt{A^2+B^2+C^2}}$.
• Between two planes with normals $(A_1,B_1,C_1)$ and $(A_2,B_2,C_2)$: $\cos\theta = \dfrac{|A_1A_2+B_1B_2+C_1C_2|}{\sqrt{A_1^2+B_1^2+C_1^2}\sqrt{A_2^2+B_2^2+C_2^2}}$.

Skew Lines and Shortest Distance

Two lines are skew when they are neither parallel nor intersecting (non-coplanar). The shortest distance formula uses the scalar triple product in the numerator and the cross-product magnitude in the denominator. CUET-style MCQs typically give the two lines in symmetric form and ask for this distance directly.

Distance Formulas

• Point to plane: $d = \dfrac{|Ax_0+By_0+Cz_0+D|}{\sqrt{A^2+B^2+C^2}}$.
• Two parallel planes $Ax+By+Cz+D_1=0$ and $Ax+By+Cz+D_2=0$: $d = \dfrac{|D_1-D_2|}{\sqrt{A^2+B^2+C^2}}$.

Typical CUET Question Patterns

• Convert direction ratios into direction cosines and verify $l^2+m^2+n^2=1$.
• Find the angle between two given lines using the dot product of their direction cosines.
• Write the equation of a plane through three points using the determinant form.
• Compute the shortest distance between two skew lines using the triple-product formula.

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🔴 Extended — Deep Study (3mo+)

Comprehensive coverage for students on a longer study timeline.

Projection and Projection Lengths

The projection of a line segment joining $P(x_1, y_1, z_1)$ and $Q(x_2, y_2, z_2)$ onto a line with direction cosines $(l, m, n)$ is

$$\text{length} = l(x_2-x_1) + m(y_2-y_1) + n(z_2-z_1).$$

This is the dot product $\vec{PQ}\cdot(l,m,n)$ and is signed; the absolute value gives the magnitude. CUET occasionally frames this as “length of projection of a vector onto another vector.”

Coplanarity of Four Points

Four points $A, B, C, D$ are coplanar iff the scalar triple product

$$(\vec{AB}\times\vec{AC})\cdot\vec{AD} = 0,$$

equivalently the determinant $\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \ x_3-x_1 & y_3-y_1 & z_3-z_1 \ x_4-x_1 & y_4-y_1 & z_4-z_1 \end{vmatrix} = 0$. This same determinant, set equal to zero, gives the equation of the plane through the first three points when $(x_4, y_4, z_4) = (x, y, z)$.

Edge Cases and Traps

• When a direction ratio is zero, the symmetric form $\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$ is not valid because division by zero is undefined. Use the vector form or write the line in two symmetric equations involving only the non-zero denominators.
• A line parallel to a plane satisfies $aA + bB + cC = 0$; a line lying in the plane additionally passes through a point of the plane.
• The shortest-distance formula degenerates (denominator = 0) when the lines are parallel, not when they intersect. Many students wrongly assume intersection makes the denominator vanish.
• Sign of $d$ in the plane equation $Ax+By+Cz+D=0$ matters when judging which side of the plane a point lies on — keep the absolute value for distance calculations.

Connection to Adjacent Topics

3D Geometry is the natural extension of 2D coordinate geometry and shares machinery with vector algebra (dot, cross, triple product), matrices and determinants (plane through three points), and analytic geometry of conics in 3D-projections context. Mastery of cross-product signs and triple-product expansion is essential before tackling skew-line and plane problems.

Worked Micro-Example

Find the shortest distance between the skew lines

$$\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4} \quad \text{and} \quad \frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}.$$

Here $\vec{a_1}-\vec{a_2} = (-1,-1,-1)$, $\vec{b_1}=(2,3,4)$, $\vec{b_2}=(3,4,5)$. Compute $\vec{b_1}\times\vec{b_2} = (-1, 2, -1)$, whose magnitude is $\sqrt{6}$. The scalar triple product $(\vec{b_1}\times\vec{b_2})\cdot(\vec{a_1}-\vec{a_2}) = (-1)(-1)+(2)(-1)+(-1)(-1) = 0$. So the shortest distance is $0$ — the lines actually intersect, despite appearances.

Common Mistakes

• Using direction ratios where the formula requires direction cosines.
• Dropping the absolute value in the point-to-plane distance.
• Forgetting to check parallelism before applying the skew-line formula.

Practice Prompts

1. A line makes angles $\alpha, \beta, \gamma$ with the coordinate axes such that $\sin^2\alpha + \sin^2\beta + \sin^2\gamma = 2$. Find the value of $\cos^2\alpha + \cos^2\beta + \cos^2\gamma$ using $l^2+m^2+n^2=1$.
2. Find the equation of the plane passing through $(1,1,1)$, $(2,3,4)$ and parallel to the z-axis, then compute the perpendicular distance of the origin from this plane.

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