Question

Let $\stackrel{\rightarrow}{a}\$ = $\stackrel{\wedge}{i}+\stackrel{\wedge}{j} +\stackrel{\wedge}{k}$, $\stackrel{\rightarrow}{b}\$ = $4\stackrel{\wedge}{i}-2\stackrel{\wedge}{j} +3\stackrel{\wedge}{k}$ , and $\stackrel{\rightarrow}{c}\$ = $\stackrel{\wedge}{i}-2\stackrel{\wedge}{j} +\stackrel{\wedge}{k}$ Find a vector of magnitude 6 units,which is parallel to the vector $2\stackrel{\rightarrow}{a}-\stackrel{\rightarrow}{b}+3\stackrel{\rightarrow}{c}$.

Using properties of determinants, prove that : $\begin{vmatrix}a+b+2c & a & b \\ c & b+c+2a & b \\ c & a & c+a+2b\end{vmatrix}$ = $2(a+b+c)^{3}.$

Form the differential equation of the family of circles in the second quadrant and touching the co-ordinate axes.

Solve the differential equation ${\frac{d y}{d x}}\ +\ y\ \cot\ x$= 2 cos x, given that y = 0 when $x={\frac{\pi}{3}}\,.$

Solve the following differential equation : ${\frac{d y}{d x}}={\frac{y}{x}}+{\frac{\sqrt{x^{2}+y^{2}}}{x}},\,x>0\,.$

Find the general solution of the differential equation : $x d y-y d x={\sqrt{x^{2}+y^{2}}}d x~.$

Using properties of determinants, prove that : $\begin{vmatrix}(a+1)(a+2) & a+2 & 1 \\ (a+2)(a+3) & a+3 & 1 \\ (a+3)(a+4) & a+4 & 1\end{vmatrix}$ = -2