Revision: Three Dimensional Geometry JEE Main Three Dimensional Geometry

The position of a point in a plane is expressed by a pair of numbers, one concerning the x-axis and the other concerning the y-axis. called co-ordinates. 

• x → distance from y-axis (abscissa)

• y → distance from x-axis (ordinate)

The two mutually perpendicular number lines intersecting each other at their zeroes are called rectangular axes or coordinate axes, or axes of reference. 

If α, β and γ are the direction angles of a vector, then the cosines of these angles, i.e.

l = cos⁡α, m = cos⁡β, n = cos⁡γ 

are called the direction cosines of the vector.

If \[ P(x,y,z) \] is a point and \[ \vec{OP} \] is its position vector, then

\[ r = |\vec{OP}| = \sqrt{x^{2} + y^{2} + z^{2}} \]

and

\[ \boxed{\,\cos\alpha = \frac{x}{r}, \qquad \cos\beta = \frac{y}{r}, \qquad \cos\gamma = \frac{z}{r}\,} \]

The angles made by a vector with the positive directions of the X-axis, Y-axis and Z-axis are called direction angles of the vector, denoted by α, β, and γ.

\[ l = \cos\alpha, \quad m = \cos\beta, \quad n = \cos\gamma, \]

\[ \boxed{\,\cos^{2}\alpha + \cos^{2}\beta + \cos^{2}\gamma = 1\,} \]

If l, m, n are direction cosines of a line and if a, b, c are real numbers such that \[\frac{\mathrm{a}}{l}=\frac{\mathrm{b}}{\mathrm{m}}=\frac{\mathrm{c}}{\mathrm{n}}=\lambda,\] then a, b, c are called direction ratios of that line.

\[ \boxed{\,\frac{a}{\sqrt{a^{2} + b^{2} + c^{2}}}, \qquad \frac{b}{\sqrt{a^{2} + b^{2} + c^{2}}}, \qquad \frac{c}{\sqrt{a^{2} + b^{2} + c^{2}}}\,} \]

The cosines of these angles are called the direction cosines of the line.

So, the direction cosines are written as (l, m, n).

\[ \boxed{\,l^{2} + m^{2} + n^{2} = 1\,} \]

If a directed line makes angles \[\alpha\], \[\beta\], and \[\gamma\] with the positive x-, y-, and z-axes respectively, then these are called the direction angles of the line.

Any three numbers proportional to the direction cosines of a line are called the direction ratios of the line.

If (a, b, c) are direction ratios, then:

An equation of the form ax + by + c = 0 represents a straight line and is known as a linear equation.

The slope m of a line is m = tan⁡θ

where θ is the inclination of the line with the positive x-axis.

\[P\left(\frac{m_1x_2+m_2x_1}{m_1+m_2},\frac{m_1y_2+m_2y_1}{m_1+m_2}\right)\]

If slopes are m₁ and m₂​:

\[\tan\theta=\left|\frac{m_1-m_2}{1+m_1m_2}\right|\]

If lines are a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0,

\[\tan\theta=\left|\frac{a_1b_2-a_2b_1}{a_1a_2+b_1b_2}\right|\]

Conditions for Parallel, Perpendicular and Identical Lines:

Parallel Lines:

Slope: m₁ = m₂

In general form: \[\frac{a_1}{a_2}=\frac{b_1}{b_2}\neq\frac{c_1}{c_2}\]

Perpendicular Lines:

Slope: m₁m₂ = −1

In general form: a₁a₂ + b₁b₂ = 0

Identical Lines:

\[\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}\]

For a₁x + b₁y + c₁ = 0 and a₂x + b₂y + c₂ = 0,

\[(x,y)=\left(\frac{b_1c_2-b_2c_1}{a_1b_2-a_2b_1},\frac{c_1a_2-c_2a_1}{a_1b_2-a_2b_1}\right)\]

\[SD=\left|\frac{\left(a_{2}-a_{1}\right)\times b}{\left|b\right|}\right|\]

If the lines are \[ \vec{r} = \vec{a}_{1} + \lambda\vec{b}_{1} \] and \[ \vec{r} = \vec{a}_{2} + \mu\vec{b}_{2}, \]

then \[ \vec{a}_{1}, \vec{a}_{2} \] are position vectors of points on the lines and \[ \vec{b}_{1}, \vec{b}_{2} \] are their direction vectors.

Vector Form: 

\[\mathbf{d}=\left|\frac{(\overline{\mathbf{b}}_{1}\times\overline{\mathbf{b}}_{2}).(\overline{\mathbf{a}}_{2}-\overline{\mathbf{a}}_{1})}{\left|\overline{\mathbf{b}}_{1}\times\overline{\mathbf{b}}_{2}\right|}\right|\]

Cartesian Form:

\[\mathbf{d}=\left|\frac{ \begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ \mathbf{a}_1 & \mathbf{b}_1 & \mathbf{c}_1 \\ \mathbf{a}_2 & \mathbf{b}_2 & \mathbf{c}_2 \end{vmatrix}}{\sqrt{\left(\mathbf{a}_1\mathbf{b}_2-\mathbf{a}_2\mathbf{b}_1\right)^2+\left(\mathbf{a}_1\mathbf{c}_2-\mathbf{a}_2\mathbf{c}_1\right)^2+\left(\mathbf{b}_1\mathbf{c}_2-\mathbf{b}_2\mathbf{c}_1\right)^2}}\right|\]

From general form:

• Slope (m) = −a / b
• Y-intercept = −c / b

If the direction ratios are \[(a_1, b_1, c_1)\] and \[(a_2, b_2, c_2)\], then:

If the direction cosines of the two lines are

\[(l_1, m_1, n_1)\] and \[(l_2, m_2, n_2)\], then:

If the direction ratios of two lines are:

First line: \[(a_1, b_1, c_1)\]

Second line: \[(a_2, b_2, c_2)\]

then the cosine of the angle \[\theta\] between them is:

\[m=\frac{y_2-y_1}{x_2-x_1}\]

For point (x₁, y₁) and line ax + by + c = 0,

\[p=\frac{|ax_1+by_1+c|}{\sqrt{a^2+b^2}}\]

For lines ax + by + c₁ = 0 and ax + by + c₂ = 0,

P = \[\left|\frac{c_{1}-c_{2}}{\sqrt{a^{2}+b^{2}}}\right|\]

If ax² + 2hxy + by² + 2gx + 2fy + c = 0 represents a pair of parallel straight lines, then the distance between them is given by

\[2\sqrt{\frac{g^{2}-ac}{a(a+b)}}\mathrm{or}2\sqrt{\frac{f^{2}-bc}{b(a+b)}}\]

\[SD=\left|\frac{\left(a_{2}-a_{1}\right)\times b}{\left|b\right|}\right|\]

If the lines are \[ \vec{r} = \vec{a}_{1} + \lambda\vec{b}_{1} \] and \[ \vec{r} = \vec{a}_{2} + \mu\vec{b}_{2}, \]

then \[ \vec{a}_{1}, \vec{a}_{2} \] are position vectors of points on the lines and \[ \vec{b}_{1}, \vec{b}_{2} \] are their direction vectors.

Vector Form: 

\[\mathbf{d}=\left|\frac{(\overline{\mathbf{b}}_{1}\times\overline{\mathbf{b}}_{2}).(\overline{\mathbf{a}}_{2}-\overline{\mathbf{a}}_{1})}{\left|\overline{\mathbf{b}}_{1}\times\overline{\mathbf{b}}_{2}\right|}\right|\]

Cartesian Form:

\[\mathbf{d}=\left|\frac{ \begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ \mathbf{a}_1 & \mathbf{b}_1 & \mathbf{c}_1 \\ \mathbf{a}_2 & \mathbf{b}_2 & \mathbf{c}_2 \end{vmatrix}}{\sqrt{\left(\mathbf{a}_1\mathbf{b}_2-\mathbf{a}_2\mathbf{b}_1\right)^2+\left(\mathbf{a}_1\mathbf{c}_2-\mathbf{a}_2\mathbf{c}_1\right)^2+\left(\mathbf{b}_1\mathbf{c}_2-\mathbf{b}_2\mathbf{c}_1\right)^2}}\right|\]

Vector Form:

\[\sin\theta=\left|\frac{\overline{\mathbf{b}}.\overline{\mathbf{n}}}{\left|\overline{\mathbf{b}}\right|.\left|\overline{\mathbf{n}}\right|}\right|\]

Cartesian Form:

\[\mathrm{sin}\theta=\frac{\mathrm{aa}_{1}+\mathrm{bb}_{1}+\mathrm{cc}_{1}}{\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}}\sqrt{\mathrm{a}_{1}^{2}+\mathrm{b}_{1}^{2}+\mathrm{c}_{1}^{2}}}\]

Vector Form:

\[\cos\theta=\left|\frac{\overline{\mathbf{n₁}}.\overline{\mathbf{n₂}}}{\left|\overline{\mathbf{n₁}}\right|.\left|\overline{\mathbf{n₂}}\right|}\right|\]

Cartesian Form:

\[\cos\theta=\left|\frac{\mathrm{a}_{1}\mathrm{a}_{2}+\mathrm{b}_{1}\mathrm{b}_{2}+\mathrm{c}_{1}\mathrm{c}_{2}}{\sqrt{\mathrm{a}_{1}^{2}+\mathrm{b}_{1}^{2}+\mathrm{c}_{1}^{2}}\sqrt{\mathrm{a}_{2}^{2}+\mathrm{b}_{2}^{2}+\mathrm{c}_{2}^{2}}}\right|\]

As the point R divides the line segment AB externally, we have either A-B-R or R-A-B.

Assume that A-B-R and `bar(AR) : bar(BR)` = m : n

∴ `(AR)/(BR) = m/n` so n(AR) = m(BR) 

As `n(bar(AR))` and `m(bar(BR))` have same magnitude and direction,

∴ `n(bar(AR)) = m(bar(BR))`

∴ `n(barr - bara) = m(barr - barb)`

∴ `nbarr - nbara = mbarr - mbarb`

∴ `mbarr - nbarr = mbarb - nbara`

∴ `(m - n)barr = mbarb - nbara`

∴ `barr = (mbarb - nbara)/(m - n)`

Hence proved.

R is a point on the line segment AB(A – R – B) and `bar(AR)` and `bar(RB)` are in the same direction.

Point R divides AB internally in the ratio m : n

∴ `(AR)/(RB) = m/n`

∴ n(AR) = m(RB)

As `n(bar(AR))` and `m(bar(RB))` have same direction and magnitude,

`n(bar(AR)) = m(bar(RB))`

∴ `n(bar(OR) - bar(OA)) = m(bar(OB) - bar(OR))`

∴ `n(vecr - veca) = m(vecb - vecr)`

∴ `nvecr - nveca = mvecb - mvecr`

∴ `mvecr + nvecr = mvecb + nveca`

∴ `(m + n)vecr = mvecb + nveca`

∴ `vecr = (mvecb + nveca)/(m + n)`

Let A, B and C be vertices of a triangle.

Let D, E and F be the mid-points of the sides BC, AC and AB respectively.

Let `bara, barb, barc, bard, bare` and `barf` be position vectors of points A, B, C, D, E and F respectively.

Therefore, by mid-point formula,

∴ `bard = (barb + barc)/2, bare = (bara + barc)/2` and `barf = (bara + barb)/2`

∴ `2bard = barb + barc, 2bare = bara + barc` and `2barf = bara + barb`

∴ `2bard + bara = bara + barb + barc`, similarly `2bare + barb = 2barf + barc = bara + barb + barc`

∴ `(2bard + bara)/3 = (2bare + barb)/3 = (2barf + barc)/3 = (bara + barb + barc)/3 = barg`  ...(Say)

Then we have `barg = (bara + barb + barc)/3 = ((2)bard + (1)bara)/(2 + 1) = ((2)bare + (1)barb)/(2 + 1) = ((2)barf + (1)barc)/(2 + 1)`

If G is the point whose position vector is `barg`, then from the above equation it is clear that the point G lies on the medians AD, BE, CF and it divides each of the medians AD, BE, CF internally in the ratio 2 : 1.

Therefore, three medians are concurrent.

Let `bara, barb, barc, bard, bare, barf` be the position vectors of the points A, B, C, D, E, F respectively.

Since D, E, F are the midpoints of BC, CA, AB respectively, by the midpoint formula

`bard = (barb + barc)/2, bare = (barc + bara)/2, barf = (bara + barb)/2`

∴ `bar(AD) + bar(BE) + bar(CF) = (bard - bara) + (bare - barb) + (barf - barc)`

= `((barb + barc)/2 - bara) + ((barc + bara)/2 - barb) + ((bara + barb)/2 - barc)`

= `1/2barb + 1/2barc - bara + 1/2barc + 1/2bara - barb + 1/2bara + 1/2barb - barc`

= `1/2(barb + barc - 2bara + bar c + bara - 2barb + bara + barb - 2barc)`

= `(bara + barb + barc) - (bara + barb + barc) = bar0`. 

Sign Convention

• Right of y-axis → +x

• Left of y-axis → −x

• Above x-axis → +y

• Below x-axis → −y

Standard Line Results

• x = 0 → y-axis

• y = 0 → x-axis

• x = a → line parallel to the y-axis

• y = b → line parallel to the x-axis

Quadrant Reminder

• Direction angles are the angles a line makes with the positive coordinate axes.

• Direction cosines are \[\cos \alpha\], \[\cos \beta\], and \[\cos \gamma\].

• If direction cosines are (l, m, n), then \[l^2 + m^2 + n^2 = 1\].

• Direction ratios are any numbers proportional to direction cosines.

• If direction ratios are (a, b, c), then corresponding direction cosines are:

• For points \[A(x_1, y_1, z_1)\], \[B(x_2, y_2, z_2)\], direction ratios of AB are \[(x_2 - x_1, y_2 - y_1, z_2 - z_1)\].

• Angle between two lines can be found using either direction cosines or direction ratios.

• Direction Cosines (DCs) of a line: \[(l, m, n) = (\cos \alpha, \cos \beta, \cos \gamma)\].

• Main identity: \[l^2 + m^2 + n^2 = 1\].

• Direction Ratios (DRs): Are proportional to DCs.

• Relation between DRs and DCs: If DRs are (a, b, c), then DCs are proportional to (a, b, c) divided by \[\sqrt{a^2 + b^2 + c^2}\].

• DRs for two points: For points \[P(x_1, y_1, z_1)\] and \[Q(x_2, y_2, z_2)\], DRs are \[(x_2 - x_1, y_2 - y_1, z_2 - z_1)\].

• Intersecting lines: SD = 0

• Parallel lines: \[SD = \frac{|(\vec{a}_2 - \vec{a}_1) \times \vec{b}|}{|\vec{b}|}\]

• Skew lines: \[SD = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}\]

Position of a Point:

For line: ax₁ + by₁ + c

• If ax₁ + by₁ + c = 0 → Point lies on the line
• If ax₁ + by₁ + c < 0 → Point lies on one side (origin side)
• If ax₁ + by₁ + c > 0 → Point lies on other side

Vector Form:

Condition for coplanarity of two lines:

Two lines r = a₁ + λb₁ and r = a₂ + μb₂ are coplanar if

(a₁ − a₂) · (b₁ × b₂) = 0

Equation of the plane containing both lines:

\[\left(\overline{\mathbf{r}}-\overline{\mathbf{a}_1}\right).\left(\overline{\mathbf{b}_1}\times\overline{\mathbf{b}_2}\right)=\mathbf{0}\] or \[\left(\overline{\mathbf{r}}-\overline{\mathbf{a}_2}\right).\left(\overline{\mathbf{b}_1}\times\overline{\mathbf{b}_2}\right)=\mathbf{0}\]

Cartesian Form:

\[\begin{vmatrix} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ \mathbf{a}_1 & \mathbf{b}_1 & \mathbf{c}_1 \\ \mathbf{a}_2 & \mathbf{b}_2 & \mathbf{c}_2 \end{vmatrix}=0\]

• The angle between two lines depends only on their directions.

• If lines do not pass through the origin, imagine parallel lines through the origin.

• The dot-product formula is the main method for solving these questions.

• In symmetric form, denominators give direction ratios.

• Zero dot product means perpendicular lines.

• Proportional direction ratios mean parallel lines.

• The required angle is generally the acute angle.

Nature of Slope

• m > 0 → rising line

• m < 0 → falling line

• m = 0 → horizontal line

• m = ∞→ vertical line

Parallel Lines 

Two lines are parallel ⇔ , their slopes are equal, m₁ = m₂

​Perpendicular Lines

Two lines are perpendicular ⇔ m₁ × m₂ = −1

​Collinearity of Three Points

Points A, B, and C are collinear

Method 1: Distance method

AB + BC = AC

Method 2: Slope method

Slope of AB = Slope of BC

• Intersecting lines: SD = 0

• Parallel lines: \[SD = \frac{|(\vec{a}_2 - \vec{a}_1) \times \vec{b}|}{|\vec{b}|}\]

• Skew lines: \[SD = \frac{|(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2)|}{|\vec{b}_1 \times \vec{b}_2|}\]

Equation of a Plane in Intercept form:

\[\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1\]

Distance of the Plane from Origin is

\[d=\frac{1}{\sqrt{\frac{1}{a^{2}}+\frac{1}{b^{2}}+\frac{1}{c^{2}}}}\]

Condition for Perpendicularity:

\[\overline{\mathbf{b}}=\lambda\overline{\mathbf{n}}\], λ is a parameter

\[\frac{\mathbf{a}_{1}}{\mathbf{a}}=\frac{\mathbf{b}_{1}}{\mathbf{b}}=\frac{\mathbf{c}_{1}}{\mathbf{c}}\]

Condition for Parallelism:

The line is parallel to the plane, if

\[\overline{\mathbf{b}}.\overline{\mathbf{n}}=0\]

aa₁ + bb₁ + cc₁ = 0