The distance between the point with position vector $-\hat{i}-5 \hat{j}-10 \hat{k}$ and the point of intersection of the line $\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}$ with the plane $x-y+z=5$, is $\_\_\_\_$ (units)
13
17
16
14
The equation of the line, where length of the perpendicular segment from the origin to the line is 4 and the inclination of the perpendicular segment with the positive direction of $X$-axis is $30^{\circ}$.
$y+\sqrt{3} x=8$
$y-\sqrt{3} x=8$
$y+\sqrt{2 x}=4 \sqrt{3}$
$y-\sqrt{2 x}=4 \sqrt{3}$
In a $\triangle A E X, T$ is the mid-point of $X E$ and $P$ is the mid-point of $E T$. If the $\triangle A P E$ is equilateral of side length equal to unity, then which of the following alternative is not correct?
$A X=\sqrt{13}$
$\angle E A T=90^{\circ}$
$\cos \triangle X A E=\frac{-1}{\sqrt{13}}$
$A T=\frac{1}{\sqrt{3}}$
Let $f$ be a non-negative function defined on the interval $[0,1]$. If $\int_0^x \sqrt{1-(f(t))^2} d t =\int_0^x f(t) d t, 0 \leq x \leq 1$ and $f(0)=0$, then
$f\left(\frac{1}{2}\right)<\frac{1}{2}$ and $f\left(\frac{1}{3}\right)>\frac{1}{3}$
$f\left(\frac{1}{2}\right)<\frac{1}{2}$ and $f\left(\frac{1}{3}\right)<\frac{1}{3}$
$f\left(\frac{1}{2}\right)>\frac{1}{2}$ and $f\left(\frac{1}{3}\right)>\frac{1}{3}$
$f\left(\frac{1}{2}\right)>\frac{1}{2}$ and $f\left(\frac{1}{3}\right)<\frac{1}{3}$
VITEEE Papers
All year-wise previous year question papers