KCET 2019

Foot of the perpendicular drawn from the point \((1,3,4)\) to the plane \(2 x-y+z+3=0\) is

Acute angel between the line \(\frac{(x-5)}{2}=\frac{y+1}{-1}=\frac{z+4}{1}\) and the plane \(3 x-4 y-z+5=0\) is

\(X Y\) plane divides the line joining the points \(A(2,3,-5)\) and \(B(-1,-2,-3)\) in the ratio

The distance of the point \((1,2,1)\) from the line \(\frac{x-1}{2}=\frac{y-2}{1}=\frac{z-3}{2}\) is.

If \(\cos x=|\sin x|\) then, the general solution is

\(\sqrt{3} \operatorname{cosec} 20^{\circ}-\sec 20^{\circ}=\)

If \(|\mathbf{a}|=16,|\mathbf{b}|=4\), then \(\sqrt{|\mathbf{a} \times \mathbf{b}|^2+|\mathbf{a} \cdot \mathbf{b}|^2}=\)

If the angle between \(\mathbf{a}\) & \(\mathbf{b}\) is \(\frac{2 \pi}{3}\) and the projection of \(\mathbf{a}\) in the direction of \(\mathbf{b}\) is \(-\)2 , the \(|\mathbf{a}|=\)

A unit vector perpendicular to the plane containing the vector \(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}\) and \(-2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}\) is

\([\mathbf{a}+2 \mathbf{b}-\mathbf{c}, \mathbf{a}-\mathbf{b}, \mathbf{a}-\mathbf{b}-\mathbf{c}]=\)