Question

You are given the line $l:\mathbf{r}=(1+2\lambda,-1-\lambda,3\lambda)$ , where $\lambda$ is a parameter.

Which of the following statements are true?

1. The point (3,-2,3) lies on the line $l$
2. The vector (2,−1,3) is parallel to the line.
3. Line $l$ is perpendicular to the line $\mathbf{r}=(2t, 2-t,-1+3t)$ , where $t$ is a parameter.
4. The equation of a plane containing line $l$ and also passing through the point $(0,2,−1)$ is $8x+y-5z=7$ .
5. The vector $(2,−1,3)$ is perpendicular to the line $l$ .

Answer 1

1.If the point (3, -2, 3) lies on the line $l: r=(1+2\lambda,-1-\lambda,3\lambda)$, then $(1+2\lambda=3)\land(-1-\lambda=-2)\land(3\lambda =3)$

All three conditions is true when $\lambda= 1$, so the point lies on the line $l$ when $\lambda = 1$

2.If the vector (2, -1, 3) is parallel to the line, then $\exist t\in R:t*(a, b, c) = (2, -1, 3)$, where a, b, c - coefficients before $\lambda$ for every coordinate

In our case: a = 2, b = -1, c = 3. So If the vector (2, -1, 3) is parallel to the line, then $\exist t\in R:t*(2, -1, 3) = (2, -1, 3)$. We have equility when t = 1, so vector (2, -1, 3) is parallel to the line

3.If two lines are perpendicular, the $a{\scriptscriptstyle 1}*a{\scriptscriptstyle 2}+b{\scriptscriptstyle 1}*b{\scriptscriptstyle 2}+c{\scriptscriptstyle 1}*c{\scriptscriptstyle 2} = 0$, where $a{\scriptscriptstyle 1},b{\scriptscriptstyle 1},c{\scriptscriptstyle 1}$ and $a{\scriptscriptstyle 2}, b{\scriptscriptstyle 2}, c{\scriptscriptstyle 2}$ are coefficients before parameters for every coordinate for first and second line

In our case: $a{\scriptscriptstyle 1}*a{\scriptscriptstyle 2}+b{\scriptscriptstyle 1}*b{\scriptscriptstyle 2}+c{\scriptscriptstyle 1}*c{\scriptscriptstyle 2} = 2*2 + (-1)*(-1)+3*3 = 14 \not = 0\to$ lines aren't perpendicular

4.Firstly we find two points that belong to the line l

$M{\scriptscriptstyle 1} = (1, -1, 0)$, when $\lambda = 0$

$M{\scriptscriptstyle 2} = (3, -2, 3)$, when $\lambda =1$

$M{\scriptscriptstyle 3} = (0, 2, -1)$

Now we have to find the equation of the plane that contains 3 points

det$\begin{pmatrix} x-1 & y+1 & z \\ 3-1 & -2+1&3 \\ 0-1&2+1&-1 \end{pmatrix}$ = (x-1)*(-2+1)*(-1)+(y+1)*3*(-1)+z*(3-1)*(2+1)-(-1)*(-2+1)*z-(2+1)*3*(x--1)-(-1)*(3-1)(y+1) = x-1-3y-3+6z-z-9x+9+2y+2 = -8x - y +5z +7 = 0

The equation of the plane is $-8x - y +5z +7 = 0 \to 8x+y-5z+7 \to$ The statement is true

5.In the second statement we found out that that vector is parallel to the line, so it's not perpendicular

So, the statements 1, 2, 4 are true