The point of intersection of the lines repres | vedclass

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

Question

(D) To find the point of intersection,we equate the two line equations: $(1+t) \overline{i} + (-6+2t) \overline{j} + (2+t) \overline{k} = (2s) \overli

Vedclass · Accepted Answer

$8 \overline{i}+8 \overline{j}+9 \overline{k}$

Vedclass · Answer

(D) To find the point of intersection,we equate the two line equations: $(1+t) \overline{i} + (-6+2t) \overline{j} + (2+t) \overline{k} = (2s) \overline{i} + (4+s) \overline{j} + (1+2s) \overline{k}$ Comparing the components,we get: $1) 1+t = 2s$ $2) -6+2t = 4+s \implies 2t-s = 10$ $3) 2+t = 1+2s \implies t-2s = -1$ From equation $(1)$,$t = 2s-1$. Substituting this into equation $(2)$: $2(2s-1) - s = 10 \implies 4s-2-s = 10 \implies 3s = 12 \implies s = 4$. Now,find $t$: $t = 2(4)-1 = 7$. Check with equation $(3)$: $7 - 2(4) = 7-8 = -1$. This is consistent. Substitute $s=4$ into the second line equation: $\overline{r} = (0 \overline{i} + 4 \overline{j} + 1 \overline{k}) + 4(2 \overline{i} + 1 \overline{j} + 2 \overline{k}) = 8 \overline{i} + 8 \overline{j} + 9 \overline{k}$.

The point of intersection of the lines represented by $\overline{r}=(\overline{i}-6 \overline{j}+2 \overline{k})+t(\overline{i}+2 \overline{j}+\overline{k})$ and $\overline{r}=(4 \overline{j}+\overline{k})+s(2 \overline{i}+\overline{j}+2 \overline{k})$ is

Similar Questions

The angle between the lines $x-3y-4=0, 4y-z+5=0$ and $x+3y-11=0, 2y-z+6=0$ is

The Cartesian equation of a line is $2x - 2 = 3y + 1 = 6z - 2$. The vector equation of the line is:

The angle between the lines $\frac{x-1}{4}=\frac{y-3}{1}=\frac{z}{8}$ and $\frac{x-2}{2}=\frac{y+1}{2}=\frac{z-4}{1}$ is

The angle between the lines whose direction cosines satisfy the equations $l+m+n=0$ and $l^2+m^2-n^2=0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module