The lines bar{r} = (hat{i} + hat{j} - hat{k}) | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The given lines are in the form $\bar{r} = \bar{a}_1 + \lambda \bar{b}_1$ and $\bar{r} = \bar{a}_2 + \mu \bar{b}_2$. Here,$\bar{b}_1 = 3 \hat{i} -

Vedclass · Accepted Answer

intersecting but not perpendicular

Vedclass · Answer

(A) The given lines are in the form $\bar{r} = \bar{a}_1 + \lambda \bar{b}_1$ and $\bar{r} = \bar{a}_2 + \mu \bar{b}_2$. Here,$\bar{b}_1 = 3 \hat{i} - \hat{j}$ and $\bar{b}_2 = 2 \hat{i} + 3 \hat{k}$. First,check if the lines are parallel: $\bar{b}_1$ is not a scalar multiple of $\bar{b}_2$,so they are not parallel. Next,check for intersection by equating the lines: $(\hat{i} + \hat{j} - \hat{k}) + \lambda(3 \hat{i} - \hat{j}) = (4 \hat{i} - \hat{k}) + \mu(2 \hat{i} + 3 \hat{k})$. Equating components: $1 + 3\lambda = 4 + 2\mu \implies 3\lambda - 2\mu = 3$ $1 - \lambda = 0 \implies \lambda = 1$ $-1 = -1 + 3\mu \implies 3\mu = 0 \implies \mu = 0$ Substituting $\lambda = 1$ and $\mu = 0$ into the first equation: $3(1) - 2(0) = 3$,which is $3 = 3$. Since the equations are consistent,the lines intersect. Check for perpendicularity: $\bar{b}_1 \cdot \bar{b}_2 = (3)(2) + (-1)(0) + (0)(3) = 6 
eq 0$. Thus,the lines are intersecting but not perpendicular.

The lines $\bar{r} = (\hat{i} + \hat{j} - \hat{k}) + \lambda(3 \hat{i} - \hat{j})$ and $\bar{r} = (4 \hat{i} - \hat{k}) + \mu(2 \hat{i} + 3 \hat{k})$ are

Similar Questions

The distance of the point $A(7, -2, 11)$ from the line $\frac{x-6}{1} = \frac{y-4}{0} = \frac{z-8}{3}$ measured along the line $\frac{x-7}{2} = \frac{y+2}{-3} = \frac{z-11}{6}$ is:

Consider a line $L$ passing through the points $P(1, 2, 1)$ and $Q(2, 1, -1)$. If the mirror image of the point $A(2, 2, 2)$ in the line $L$ is $(\alpha, \beta, \gamma)$,then $\alpha + \beta + 6\gamma$ is equal to:

If the direction ratios of two lines are given by $3lm - 4ln + mn = 0$ and $l + 2m + 3n = 0$,then the angle between the lines is

The vector equation of the line passing through $P(1, 2, 3)$ and $Q(2, 3, 4)$ is

If the lines $\frac{x - 1}{-3} = \frac{y - 2}{2k} = \frac{z - 3}{2}$ and $\frac{x - 1}{3k} = \frac{y - 5}{1} = \frac{z - 6}{-5}$ are perpendicular to each other,then $k = \dots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module