COMEDK 2024 Morning Shift

Find the direction in which a straight line must be drawn through the point \((1,2)\) so that its point of intersection with the line \(x+y=4\) may be at a distance of \(\sqrt{\frac{2}{3}}\) from this point.

The perpendicular distance of a line from the origin is 5 units and its slope is \(-1\).
The equation of the line is

The vector equation of two lines are
$$\begin{aligned} & \vec{r}=(1-t) \hat{\imath}+(t-2) \hat{\jmath}+(3-2 t) \hat{k} \\ & \vec{r}=(s+1) \hat{\imath}+(2 s-1) \hat{\jmath}-(2 s+1) \hat{k} \end{aligned}$$
Then the shortest distance between t...

\(P\) is a point on the line segment joining the points \((3,2,-1)\) and \((6,2,-2)\). If \(x\) coordinate of \(\mathrm{P}\) is 5, then its \(y\) co-ordinate is

The foot of the perpendicular from \((2,4,-1)\) to the line \(x+5=\frac{1}{4}(y+3)=-\frac{1}{9}(z-6)\) is

If \(\cos \theta=\frac{1}{2}\left(x+\frac{1}{x}\right)\) then \(\frac{1}{2}\left(x^2+\frac{1}{x^2}\right)=\)

\(4\left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{5 \pi}{8}\right)\left(1+\cos \frac{7 \pi}{8}\right) \text { is equal to }\)

\(\text { If } \sin A=\frac{4}{5} \text { and } \cos B=\frac{-12}{13} \text { where } A \text { and } B \text { lie in first and third quadrant respectively. Then } \cos (A+B)=\)

\(\text { If }|\vec{a} \times \vec{b}|^2+|\vec{a} \cdot \vec{b}|^2=144 ~\&~|\vec{a}|=4 \text { then }|\vec{b}|=\)

The vector \((\vec{r})\) whose magnitude is \(3 \sqrt{2}\) units which makes an angle of \(\frac{\pi}{4}\) and \(\frac{\pi}{2}\) with \(y\) and \(z\)- axis respectively is