Question

3.Given that x + 2y + 3z =1, 3x + 2y + z = 4, x + 3y + 2z = 0. What is x,y and z?
a.7/4,−3/4,1/4
b.5/4,−2/4,¼
c.\(1,-3/4,1/4)
d.None of the option

4.What is the angle between the two linesâˆ′3x+4y=8 and âˆ′2xâˆ′8yâˆ′14=0?
a.14.63 degree
b.âˆ′14degree
c.−14.63degree
d.14 degree

Answer 1

Given that x + 2y + 3z =1, 3x + 2y + z = 4, x + 3y + 2z = 0, Cramer's rule:

$$x=\frac{\begin{Vmatrix} 1 & 2 & 3 \\ 4 & 2 & 1 \\ 0 & 3 & 2 \end{Vmatrix}} {\begin{Vmatrix} 1 & 2 & 3 \\ 3 & 2 & 1 \\ 1 & 3 & 2 \end{Vmatrix}} = \frac{21}{12}=\frac{7}{4}\,\,$$ $$y=\frac{\begin{Vmatrix} 1 & 1 & 3 \\ 3 & 4 & 1 \\ 1 & 0 & 2 \end{Vmatrix}} {\begin{Vmatrix} 1 & 2 & 3 \\ 3 & 2 & 1 \\ 1 & 3 & 2 \end{Vmatrix}} = \frac{-9}{12}=\frac{-3}{4}\,\,$$

$$z=\frac{\begin{Vmatrix} 1 & 2 & 1 \\ 3 & 2 & 4 \\ 1 & 3 & 0 \end{Vmatrix}} {\begin{Vmatrix} 1 & 2 & 3 \\ 3 & 2 & 1 \\ 1 & 3 & 2 \end{Vmatrix}} = \frac{3}{12}=\frac{1}{4}$$

Solution: a. $7/4, \, -3/4, \, 1/4$

What is the angle between the two lines $a\cdot x+b \cdot y=c$ and $d\cdot x + e\cdot y =f$ ? 

Vectors, perpendicular to lines: $\vec{v}_1 = (a,b), \, \vec{v}_2 = (d,e)$

Angle between vectors:

$$\cos{\phi}= \frac{\vec{v}_1\cdot \vec{v}_2}{|\vec{v}_1|\cdot|\vec{v}_2|} = \frac{ad+eb}{\sqrt{a^2+b^2}\sqrt{d^2+e^2}}$$

If $\cos{\phi}<0 \Rightarrow \phi> \pi/2$ , and angle between lines equal to $\pi - \phi$ .

Therefore, angle between lines:

$$\phi=\arccos \left| \frac{ad+eb}{\sqrt{a^2+b^2}\sqrt{d^2+e^2}}\right|$$