Revision: Three Dimensional Geometry JEE Main Three Dimensional Geometry

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.

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.

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 point is (x,y,z) and distance r: \[\cos\alpha=\frac{x}{r},\quad\cos\beta=\frac{y}{r},\quad\cos\gamma=\frac{z}{r}\]

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:

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.

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

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

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.

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

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

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

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:

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

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

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 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:

If the direction cosines of the two lines are

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

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

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

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

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

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

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:

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

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`. 

• 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

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