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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;  tR\underline{r}=\underline{r_0}+t\underline{v};\;t\in\mathbb{R}

Symmetric equation

xx0a=yy0b=zz0c\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=(1t)rA+trB;  tR\underline{r}=(1-t)\underline{r_A}+t\underline{r_B};\;t\in\mathbb{R}

Symmetric equation

xx1x2x1=yy1y2y1=zz1z2z1\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:xx1a1=yy1b1=zz1c1;    β:xx2a2=yy2b2=zz2c2\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,c1v_1=\langle a_1, b_1, c_1\rangle, v2=a2,b2,c2v_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×v2v1×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:

cosθ=v1v1v1v2=(a1i+b1j+c1k)(a2i+b2j+c2k)a1i+b1j+c1ka2i+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=rp2[n(rp)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