COMEDK 2024 Afternoon Shift

Let \(\mathrm{ABC}\) be a triangle with equations of its sides \(\mathrm{AB}, \mathrm{BC}\). \(\mathrm{CA}\) respectively are \(x-2=0, y-5=0\) and \(5 x+2 y-10=0\). Then the orthocentre of triangle lies on the line

For what value of \(\mathrm{a}\) and \(\mathrm{b}\) the intercepts cut off on the co-ordinate axes by the line \(a x-b y+8=0\) are equal in length but opposite in signs to those cut off by the line \(2 x-3 y+6=0\) on the axes

The lines
\(\vec{r}=(2 \hat{\jmath}-3 \hat{k})+\lambda(\hat{\imath}+2 \hat{\jmath}+3 \hat{k})\) and \(\vec{r}=(2 \hat{\imath}+6 \hat{\jmath}+3 \hat{k})+\mu(2 \hat{\imath}+3 \hat{\jmath}+4 \hat{k})\) are

A line makes the same angle \(\theta\) with each of the \(x\) and \(z\)-axes. If the angle \(\beta\), which it makes with the \(y\)-axis is such that \(\sin ^2 \beta=3 \sin ^2 \theta\), then \(\cos ^2 \theta\) equals

If the line \(\frac{1-x}{-3}=y=\frac{z+2}{2}\) is perpendicular to the line \(\frac{3 x-1}{2 b}=3-y=\frac{z-1}{a}\), then find the value of \(3 a+3 b\)

\(\left(\cos \frac{\pi}{12}-\sin \frac{\pi}{12}\right)\left(\tan \frac{\pi}{12}+\cot \frac{\pi}{12}\right)=\)

\(\frac{\cos 9^{\circ}+\sin 9^{\circ}}{\cos 9^{\circ}-\sin 9^{\circ}}=\)

\(\text { If } \frac{x}{\cos \theta}=\frac{y}{\cos \left(\theta+\frac{2 \pi}{3}\right)}=\frac{z}{\cos \left(\theta-\frac{2 \pi}{3}\right)} \text { then } x+y+z \text { is equal to }\)

Let a, b, c be three vector such that \(a \neq 0\) and \(\vec{a} \times \vec{b}=2 \vec{a} \times \vec{c},|a|=|c|=1,|b|=4\) and \(|\vec{b} \times \vec{c}|=\sqrt{15}\). If \(\vec{b}-2 \vec{c}=\lambda \vec{a}\) then \(\lambda\) equals to

\(\text { The angle between } \hat{\imath}-\hat{\jmath} ~\&~ \hat{\jmath}-\hat{k} \text { is }\)