Condition for Two Lines to Be Perpendicular

Q. Let two lines $L_1:\frac{x-3}{3}=\frac{y-2}{-4}=\frac{z}{-1}$ and $L_2:\frac{x-3}{4}=\frac{y-2}{1}=\frac{z}{-3}$ and $\frac{x-3}{a}=\frac{y-2}{6}=\frac{z}{b},(a,b\in R)$ is the acute angle bisector between lines $L_1$ and $L_2.$ Then the value of $|a+b|=$

Q. If $l_1,m_1,n_1$ and $l_2,m_2,n_2$ are the direction cosines of two perpendicular lines, then the direction cosine of the line which is perpendicular to both the lines, will be

1. $(m_1{n}_2-m_2{n}_1),(n_1{l}_2-l_1{n}_2),(l_1{m}_2-l_2{m}_1)$
2. $(l_1{l}_2-m_2{m}_1),(m_1{m}_2-n_1{n}_2),(n_1{n}_2-l_2{l}_1)$
3. $\frac{1}{\sqrt{l_1^2+m_1^2+n_1^2}},\frac{1}{\sqrt{l_2^2+m_2^2+n_2^2}},\frac{1}{\sqrt{3}}$
4. $\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}}$

Q. If the co-ordinates of the points P and Q be (1, -2, 1) and (2, 3, 4) and O be the origin, then

1. OP $\perp$ OQ
2. OP = OQ
3. None of these
4. OP || OQ

Q.

A line l passing through the origin is perpendicular to the lines
$l_1:(3+t)\hat{i}+(-1+2t)\hat{j}+(4+2t)\hat{k},-\infty <t<\infty l_2:(3+2s)\hat{i}+(3+2s)\hat{j}+(2+s)\hat{k},-\infty <s<\infty$
Then, the coordinate(s) of the point(s) on $l_2$ at a distance of $\sqrt{17}$ from the point of intersection of l and $l_1$ is (are)

1. $(\frac{7}{3},\frac{7}{3},\frac{5}{3})$

2. $-1,-1,0$

3. $(1,1,1)$

4. $(\frac{7}{9},\frac{7}{9},\frac{8}{9})$

Q.

From a point $P(\lambda ,\lambda ,\lambda )$, perpendiculars PQ and PR are drawn respectively on the lines y = x, z = 1 and y = -x, z = -1. If P is such that $\angle QPR$ is a right angle, then the possible value(s) of
$\lambda$ is (are)

1. 1

2. -1

3. $\sqrt{2}$

4. $-\sqrt{2}$

Q. $\begin{array}{cc}\text{Column I}& \text{Column II}\\ \text{a. Lines}\frac{x-1}{-2}=\frac{y+2}{3}=\frac{z}{-1}\text{and}& \\ \overrightarrow{r}=(3\hat{i}-\hat{j}+\hat{k})+t(\hat{i}+\hat{j}+\hat{k})\text{are}& \text{p. intersecting}\\ \text{b. Lines}\frac{x+5}{1}=\frac{y-3}{7}=\frac{z+3}{3}\text{and}& \\ x-y+2z-4=0=2x+y-3z+5=0& \\ \text{are}& \text{q. perpendicular}\\ \text{c. Lines}(x=t-3,y=-2t+1,z=-3t-2)& \\ \text{and}\overrightarrow{r}=(t+1)\hat{i}+(2t+3)\hat{j}+(-t-9)\hat{k}& \\ \text{are}& \text{r. parallel}\\ \text{d. Lines}\overrightarrow{r}=(\hat{i}+3\hat{j}-\hat{k})+t(2\hat{i}-\hat{j}-\hat{k})\text{and}& \\ \overrightarrow{r}=(-\hat{i}-2\hat{j}+5\hat{k})+s(\hat{i}-2\hat{j}+\frac{3}{4}\hat{k})\text{are}& \text{s. skew}\end{array}$

Then which of the following is correct ?

1. $a\to q,s;b\to r;c\to p,q;d\to p$
2. $a\to q;b\to r;c\to p,q;d\to p,q$
3. $a\to s;b\to r,s;c\to p;d\to p,q$
4. $a\to q,s;b\to r,s;c\to p;d\to p$

Q. Equation of a line in the plane $\pi :2x-y+z-4=0$ which is perpendicular to the line $l$ whose equation is $\frac{x-2}{1}=\frac{y-2}{-1}=\frac{z-3}{-2}$ and which passes through the point of intersection of $l$ and $\pi$ is

1. $\frac{x-2}{1}=\frac{y-1}{5}=\frac{z-1}{-1}$
2. $\frac{x+2}{2}=\frac{y+1}{-1}=\frac{z+1}{1}$
3. $\frac{x-2}{2}=\frac{y-1}{-1}=\frac{z-1}{1}$
4. $\frac{x-1}{3}=\frac{y-3}{5}=\frac{z-5}{-1}$

Q. If the two lines $x+(a-1)y=1$ and $2x+a^{2}y=1,(a\in R-\{0,1\left\}\right)$ are perpendicular, then the distance of their point of intersection from the origin is :

1. $\frac{\sqrt{2}}{5}$
2. $\frac{2}{\sqrt{5}}$
3. $\sqrt{\frac{2}{5}}$
4. $\frac{2}{5}$

Q. If the foot of the perpendicular drawn from the point $(1,0,3)$ on a line passing through $(\alpha ,7,1)$ is $(\frac{5}{3},\frac{7}{3},\frac{17}{3})$, then $\alpha$ is equal to

Q. If the lines $x=ay+b,z=cy+d$ and $x=a^{\prime }z+b^{\prime },y=c^{\prime }z+d^{\prime }$ are perpendicular, then :

1. $a{b}^{\prime }+b{c}^{\prime }+1=0$
2. $b{b}^{\prime }+c{c}^{\prime }+1=0$
3. $a{a}^{\prime }+c+c^{\prime }=0$
4. $c{c}^{\prime }+a+a^{\prime }=0$