If the equation of the line passing through the point (0, -1/2, 0) and perpendicular to the lines

Question

If the equation of the line passing through the point \(\left(0,-\frac{1}{2}, 0\right)\) and perpendicular to the lines \(\overrightarrow{\mathrm{r}}=\lambda(\hat{\mathrm{i}}+\mathrm{a} \hat{\mathrm{j}}+\mathrm{b} \hat{\mathrm{k}})\) and \(\overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}-\hat{\mathrm{j}}-6 \hat{\mathrm{k}})+\mu(-\mathrm{b} \hat{\mathrm{i}}+a \hat{\mathrm{j}}+5 \hat{\mathrm{k}})\) is \(\frac{x-1}{-2}=\frac{y+4}{d}=\frac{z-c}{-4},\) then \(a+b+c+d\) is equal to :

(1) 10

(2) 14

(3) 13

(4) 12  

Answer 1

Correct option is: (2) 14 

Line is \(\perp^{\mathrm{r}}\) to 2 line \(\Rightarrow\) line will be parallel to

\((i+a \hat{j}+b \hat{k}) \times(-b \hat{i}+a \hat{j}+5 \hat{k})\)

Parallel vector along the required line is

\(\hat{\mathrm{i}}(5 \mathrm{a}-\mathrm{ab})-\hat{\mathrm{j}}\left(\mathrm{b}^{2}+5\right)+\hat{\mathrm{k}}(\mathrm{a}+\mathrm{ab})\)

Dr's of required line \(\alpha(5 a-a b),-\left(b^{2}+5\right),(a+a b)\)

Also Dr's of required line \(\alpha-2, \mathrm{~d},-4\)

\(\begin{equation*} \therefore \frac{5 \mathrm{a}-\mathrm{ab}}{-2}=\frac{-\left(\mathrm{b}^{2}+5\right)}{\mathrm{d}}=\frac{\mathrm{a}+\mathrm{ab}}{-4}\quad \cdots(1) \end{equation*} \)

Also point \(\left(0, \frac{-1}{2}, 0\right)\) will lie on \(\frac{x-1}{-2}=\frac{y+4}{d}=\frac{z-c}{-4}\)

\(\frac{0-1}{-2}=\frac{\frac{-1}{2}+4}{d}=\frac{0-c}{-4} \Rightarrow d=7, c=2\)

From (1) \(\frac{5 \mathrm{a}-\mathrm{ab}}{-2}=\frac{-\mathrm{b}^{2}-5}{7}=\frac{\mathrm{a}+\mathrm{ab}}{-4}\)   

\(\frac{5 a-a b}{-2}=\frac{a+a b}{-4} ; \frac{-b^{2}-5}{7}=\frac{a+a b}{-4} \)  

\(\begin{array}{c|c|c} -20 a+4 a b=-2 a-2 a b & 4 b^{2}+20=70+7 a b \\ 36+20=70+21 a \\ 18 a=6 a b & 56=28 a \Rightarrow a=2\\ b=3 \end{array} \\ \begin{array}{cl} \end{array} \)

a + b + c + d = 2 + 3 + 2 + 7 = 14