Answer to Question #251897 in Analytic Geometry for moe

Question

You are given the following three points A(1,−3,5), B(4,-11,1) and C(-3,8,5).

Which of the following statements are true?

1. The vector equation of the line AB is $\mathbf{r}=(1,-3,5)+\lambda (3,-8,-4)$ , where $\lambda$ is a parameter.
2. The three points  A, B  and C ie on the same straight line
3. $\underline{AB} \times \underline{AC}=(44,16,1)$
4. Lines AB and AC are perpendicular.
5. The equation of a plane passing through the points A, B and C is $44x+16y+z=1$

Answer 1

 Given three points are  A(1,−3,5), B(4,-11,1) and C(-3,8,5).

Let us check every statement true or false individually.

Statement-1.

Vector equation of line AB is

$\textbf{r}$ = (1,-3,5) + $\lambda$ (4-1, -11+3,1-5)

=> $\textbf{r}$ = (1,-3,5) + $\lambda$ (3, -8,-4)

So Statement-1 is $\textbf{true}$ .

Statement-2.

The three points A, B and C will lie on the same straight line if position vector of C satisfy the equation of AB.

i.e. if (-3,8,5)=(1,-3,5)+$\lambda$ (3,-8,4)

i.e if (-3,8,5)=(1+3$\lambda$ ,-3-8$\lambda$ ,5+4$\lambda$ )

i.e if -3 = 1+3$\lambda$ , 8=-3-8$\lambda$ ,5=5+4$\lambda$

i.e if $\lambda$ = $-\frac{4}{3}$ , $\lambda$ =$-\frac{11}{8}$ , $\lambda$ =0

As values of $\lambda$ are different, position vector of C doesn't satisfy the equation of line AB.

So statement-2 is $\textbf{false}$

Statement-3.

$\overrightarrow{AB}$ = (4-1, -11+3,1-5)=(3,-8,-4)

$\overrightarrow{AC}$ = (-3-1,8+3,5-5)=(-4,11,0)

$\overrightarrow{AB}$ X$\overrightarrow{AC}$ = $\begin{vmatrix} \hat{i} &\hat{j} &\hat{k}\\ 3 & -8&-4\\-4&11&0 \end{vmatrix}$

= $\hat{i}$ (44)+$\hat{j}$ (16-0)+$\hat{k}$ (33-32) = (44, 16, 1)

So Statement-3 is $\textbf{true}$ .

Statement-4.

If line AB and line AC are perpendicular then $\overrightarrow{AB}.$ $\overrightarrow{AC}$ = 0.

$\overrightarrow{AB}$ .$\overrightarrow{AC}$ = (3,-8,-4).(-4,11,0)

=-12-88-0=-100 ≠0

So line AB is not perpendicular to line AC

So Statement-4 is $\textbf{false}$

Statement-5

As every linear equation of x, y, z represents a plane and one and only one plane can be drawn through three non collinear points, equation of plane through A,B and C will be 44x+16y+z=1

If co-ordinates of the points A, B,C satisfy the equation of plane 44x+16y+z=1 .

For point A, 44.(1)+16.(-3)+1.(5)=1

=> 44-48+5=1

=> -4+5=1

=> 1=1

So A lies on the given plane.

For point B, 44.(4)+16.(-11)+1.(1)=1

=> 176-176+1=1

=> 1=1

So B lies on the given plane

For point C, 44.(-3)+16.(8)+1.(5)=1

=> -132+128+5=1

=> 1=1

So C lies on the given plane

So equation of plane through A, B, C is 44x+16y+z=1 .

Therefore Statement-5 is $\textbf{true}$ .