Question

Find the vector equation of the line joining (1, 2, 3) and (–3, 4, 3) and show that it is perpendicular to the z-axis.

Answer 1

we know, vector equation of a line passing through two points $\vec{a}$ and $\vec{b}$ is given by, $\vec{r}=\vec{a}+\kappa(\vec{b}-\vec{a})$

here, two points are : $\vec{a}:(1,2,3)$ and $\vec{b}:(-3,4,3)$

so, $\vec{a}=\hat{i}+2\hat{j}+3\hat{k}$
and $\vec{b}=-3\hat{i}+4\hat{j}+3\hat{k}$

so, $\vec{r}=\hat{i}+2\hat{j}+3\hat{k}+\kappa\{(-3-1)\hat{i}+(4-2)\hat{j}+(3-3)\hat{k}\}$

hence, equation of line is $\vec{r}=\hat{i}+2\hat{j}+3\hat{k}+\kappa(-4\hat{i}+2\hat{j})$

also we can prove that, line $\vec{r}$ is perpendicular to the z axis by using dot product.

unit vector along z - axis is $\hat{k}$
so, direction ratio of z - axis (0, 0, 1)

and direction ratio of line : (-4, 2, 0)

we know, if two lines of direction ratios $(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$ are perpendicular to each other then, $a_1a_2+b_1b_2+c_1c_2=0$

here, 0 × -4 + 0 × 2 + 1 × 0 = 0
hence, line is perpendicular to the z - axis.

Answer 2

we know that very well that equation of vector for a line passing through 2 points vector a and vector b can be given by,  

vector a= (1 ,2, 3) & vector b= (-3 , 4 , 3)  

total vector is denoted by R,

R=a+k(b-a)==>C

vector a= 1i+2j+3k

vector b= -3i+4j+3k

put the values of vector a and vector b in equation C

vector a=a

vector b=b

R=i+2j+3k+k(-3i+4j+3k-i-2j-3k)

R=i+2j+3k+k(-4i+2j)

R is a vector for a line passing through 2 points vector a and vector b can be given by,  

R=i+2j+3k+k(-4i+2j)

now with the help of doth product,

vector called as unit vector along with z-axis is denoted by k,

direction ration is (0,0,1)

and direction line (-4 2 0)

if two lines perpendicular to each other then we know that,

=0*-4+0*2+1*0

=0

so that it is perpendicular to axis Z.