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Sahithyan's S1
Sahithyan's S1 — Mathematics

Straight Lines

Passes through a point & parallel to a vector

Equation for a line that:

  • passes through r0‾=⟨x0,y0,z0⟩\underline{r_0}=\langle x_0, y_0, z_0 \rangle
  • is parallel to v‾=ai‾+bj‾+ck‾\underline{v}=a\underline{i}+b\underline{j}+c\underline{k}

Parametric equation

r‾=r0‾+tv‾;  t∈R\underline{r}=\underline{r_0}+t\underline{v};\;t\in\mathbb{R}

Symmetric equation

x−x0a=y−y0b=z−z0c\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}

Passes through 2 points

Equation of a line passes through A=(x1,y1,z1)A=(x_1,y_1,z_1), B=(x2,y2,z2)B=(x_2,y_2,z_2). rA‾\underline{r_A} and rB‾\underline{r_B} are the position vectors of AA and BB.

Parametric equation

r‾=(1−t)rA‾+trB‾;  t∈R\underline{r}=(1-t)\underline{r_A}+t\underline{r_B};\;t\in\mathbb{R}

Symmetric equation

x−x1x2−x1=y−y1y2−y1=z−z1z2−z1\frac{x-x_1}{x_2-x_1} = \frac{y-y_1}{y_2-y_1} = \frac{z-z_1}{z_2-z_1}

Intersection

To show that two straight lines intersect in 3D space, existence of a point which satisfies both lines must be proven.

It is not enough to show that the cross product of their parallel vectors is non-zero.

Normal to 2 lines

Let α,β\alpha,\beta be two lines.

α:x−x1a1=y−y1b1=z−z1c1;    β:x−x2a2=y−y2b2=z−z2c2\alpha:\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1};\;\; \beta:\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}

Here v1=⟨a1,b1,c1⟩v_1=\langle a_1, b_1, c_1\rangle, v2=⟨a2,b2,c2⟩v_2=\langle a_2, b_2, c_2\rangle are 22 vectors parallel to α,β\alpha, \beta respectively.

Normal to both lines: v1×v2v_1 \times v_2.

Unit normal

v1×v2∣v1×v2∣\frac{v_1 \times v_2}{|v_1 \times v_2|}

Angle between 2 straight lines

Using the α,β\alpha,\beta lines mentioned above:

cos⁡θ=v1⋅v1∣v1∣⋅∣v2∣=(a1i‾+b1j‾+c1k‾)⋅(a2i‾+b2j‾+c2k‾)∣a1i‾+b1j‾+c1k‾∣⋅∣a2i‾+b2j‾+c2k‾∣\cos{\theta} = \frac{v_1\cdot v_1}{|v_1|\cdot|v_2|} = \frac{ (a_1\underline{i}+b_1\underline{j}+c_1\underline{k}) \cdot (a_2\underline{i}+b_2\underline{j}+c_2\underline{k}) }{ \lvert{a_1\underline{i}+b_1\underline{j}+c_1\underline{k}}\rvert \cdot \lvert{a_2\underline{i}+b_2\underline{j}+c_2\underline{k}}\rvert }

Here v1,v2v_1, v_2 are 22 vectors parallel to α,β\alpha, \beta respectively.

Shortest distance from a point

The distance can be calculated using Pythogoras’ theorem.

d2=∣r‾−p‾∣2−[n‾⋅(r‾−p‾)∣n‾∣]2d^2 = {\big\lvert\underline{r} - \underline{p}\big\rvert}^2 - \bigg[ \frac{\underline{n}\cdot (\underline{r} - \underline{p})}{\lvert \underline{n} \rvert} \bigg]^2

Here:

  • PP is the arbitrary point
  • p‾\underline{p} is the position vector of PP
  • r‾\underline{r} is the position vector of a point on the line
  • n‾\underline{n} is parallel to the line