$$ \text { If the projection of } \vec{a}=5 \hat{\imath}+\hat{\jmath}+\lambda \hat{k} \text { on } \vec{b}=2 \hat{\imath}+6 \hat{\jmath}+3 \hat{k} \text { is } 4 \text { units, then } \lambda= $$
4
6
3
5
The equation of the perpendicular drawn from the point $A(6,1,3)$ to the line $\frac{x-1}{2}=\frac{2-y}{-1}=\frac{z-3}{2}$ is $\frac{x-6}{\boldsymbol{a}}=\frac{y-1}{\boldsymbol{b}}=\frac{z-3}{\boldsymbol{c}}$. If $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are the possible integers such that $\boldsymbol{a}<0$, then the value of $\boldsymbol{a}-\boldsymbol{b}+\mathbf{5} \boldsymbol{c}$ is:
$0$
$11$
$5$
-$17$
A line L passes through the point of intersection of the lines $3 x+y-10=0$ and $x-y-2=0$.
If the perpendicular distance of the line $L$ from the point $(5,1)$ is exactly $\frac{2}{\sqrt{5}}$ units, which of the following represents the correct equation for line L ?
$x+2 y-5=0$
$2 x+y-7=0$
$x-2 y+1=0$
$2 x-y-5=0$
$$ \text { The particular solution of the differential equation }(x-y)(d x+d y)=(d x-d y) \text { when } y=-1 \text { and } x=0 \text { is } $$
$$ \log |x+y|=x-y+1 $$
$$ \log \left|\frac{x-y}{x+y}\right|=1 $$
$$ \log |x-y|=x+y+1 $$
$$ \log |x-y|=x-y+1 $$
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