Vectors — Maths STD 12 Science — Question

intersection. We find the points of intersection of $y=x^2 \ldots$ (1)

and $y=x^3 \ldots$ (2)

From (1) and (2)

$\begin{aligned} & x^3=x^2 \\ & \therefore x^3-x^2=0 \\ & \therefore x^2(x-1)=0 \\ & \therefore x=0 \text { or } x=1 \\ & \text { When } x=0, y=0 . \\ & \text { When } x=1, y=1 .\end{aligned}$

For $y=x^2, \frac{d y}{d x}=2 x$

For $y=x^3, \frac{d y}{d x}=3 x^2$

Angle at $\mathrm{O}=(0,0)$

Slope of tangent to $y=x^2$ at $O$

$=\left(\frac{d y}{d x}\right)_{\text {at } O(0,0)}=2 \times 0=0$

$\therefore$ equation of tangent to $y=x^2$ at $\mathrm{O}$ is $y=0$.

Slope of tangent to $y=x^3$ at $\mathrm{O}=\left(\frac{d y}{d x}\right)_{\text {at } \mathrm{O} 0,0)}=3 \times 0=0$

$\therefore$ equation of tangent to $y=x^3$ at $P$ is $y=0$.

∴ the tangents to both curves at (0, 0) are y = 0 ∴ angle between them is 0. Angle at P = (1, 1)

$\therefore$ equation of tangent to $y=x^3$ at $P$ is $y-1=3(x-1) y=3 x-2$

We have to find angle between y = 2x – 1 and y = 3x – 2 Lines through origin parallel to these tagents are y = 2x and y = 3x

$\therefore \frac{x}{1}=\frac{y}{2}$ and $\frac{z}{1}=\frac{y}{3}$

These lines lie in XY-plane. ∴ the direction ratios of these lines are 1, 2, 0 and 1, 3, 0. The angle θ between them is given by

$\cos \theta=\frac{(1)(1)+(2)(3)+(0)(0)}{\sqrt{1^2+2^2+0^2} \sqrt{1^2+3^2+0^2}}$

$=\frac{1+6+0}{\sqrt{5} \sqrt{10}}=\frac{7}{\sqrt{50}}=\frac{7}{5 \sqrt{2}}$

$\therefore \theta=\cos ^{-1}\left(\frac{7}{5 \sqrt{2}}\right)$

Hence, the required angles are 0 and $\cos ^{-1}\left(\frac{7}{5 \sqrt{2}}\right)$.