Let O be the origin,and M and N be the points | vedclass

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

Question

(B) Let the lines be $L_1: \frac{x-5}{4}=\frac{y-4}{1}=\frac{z-5}{3}=\lambda$ and $L_2: \frac{x+8}{12}=\frac{y+2}{5}=\frac{z+11}{9}=\mu$.Any point $M$

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(B) Let the lines be $L_1: \frac{x-5}{4}=\frac{y-4}{1}=\frac{z-5}{3}=\lambda$ and $L_2: \frac{x+8}{12}=\frac{y+2}{5}=\frac{z+11}{9}=\mu$.Any point $M$ on $L_1$ is $(4\lambda+5, \lambda+4, 3\lambda+5)$ and any point $N$ on $L_2$ is $(12\mu-8, 5\mu-2, 9\mu-11)$.The vector $\overrightarrow{MN} = (12\mu-4\lambda-13, 5\mu-\lambda-6, 9\mu-3\lambda-16)$.The direction vectors are $\vec{b}_1 = (4, 1, 3)$ and $\vec{b}_2 = (12, 5, 9)$.The shortest distance vector $\overrightarrow{MN}$ must be perpendicular to both $\vec{b}_1$ and $\vec{b}_2$.$\vec{b}_1 	imes \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 4 & 1 & 3 \ 12 & 5 & 9 \end{vmatrix} = -6\hat{i} + 0\hat{j} + 8\hat{k}$.Since $\overrightarrow{MN}$ is parallel to $\vec{b}_1 	imes \vec{b}_2$,we have $\frac{12\mu-4\lambda-13}{-6} = \frac{5\mu-\lambda-6}{0} = \frac{9\mu-3\lambda-16}{8}$.From the middle term,$5\mu-\lambda-6=0 \implies \lambda = 5\mu-6$.Substituting into the ratio $\frac{12\mu-4\lambda-13}{-6} = \frac{9\mu-3\lambda-16}{8}$:$8(12\mu-4(5\mu-6)-13) = -6(9\mu-3(5\mu-6)-16) \implies 8(-8\mu+11) = -6(-6\mu+2) \implies -64\mu+88 = 36\mu-12 \implies 100\mu = 100 \implies \mu=1$.Then $\lambda = 5(1)-6 = -1$.Thus,$M = (4(-1)+5, -1+4, 3(-1)+5) = (1, 3, 2)$ and $N = (12(1)-8, 5(1)-2, 9(1)-11) = (4, 3, -2)$.Finally,$\overrightarrow{OM} \cdot \overrightarrow{ON} = (1)(4) + (3)(3) + (2)(-2) = 4 + 9 - 4 = 9$.

Similar Questions

If the two lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$ have a point in common,then $k=$

The equation of the line passing through the point $(-1, 3, -2)$ and perpendicular to each of the lines $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$ and $\frac{x+2}{-3} = \frac{y-1}{2} = \frac{z+1}{5}$ is

The shortest distance between the lines $\overline{r} = (4\hat{i} - \hat{j}) + \lambda(\hat{i} + 2\hat{j} - 3\hat{k})$ and $\overline{r} = (\hat{i} - \hat{j} + 2\hat{k}) + \mu(2\hat{i} + 4\hat{j} - 5\hat{k})$ is:

If the lines $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ and $\frac{x - 3}{1} = \frac{y - c}{2} = \frac{z}{1}$ intersect, then the radius of the circle $x^2 + y^2 - 4x + 10y + c = 0$ is

The angle between the lines $\vec{r}=(2 \hat{i}-3 \hat{j}+\hat{k})+\lambda(\hat{i}+4 \hat{j}+3 \hat{k})$ and $\vec{r}=(\hat{i}-\hat{j}+2 \hat{k})+\mu(\hat{i}+2 \hat{j}-3 \hat{k})$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module