23. Find the vector equation of the line joining (1, 2, 3) and (3,4,3)

1.	23. Find the vector equation of the line joining (1, 2, 3) and (3,4,3) and show ha itis peapendicular tothe z-axis.
Answer» 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+02+1*0 =0 so that it is perpendicular to axis Z.

23. Find the vector equation of the line joining (1, 2, 3) and (3,4,3) and show ha itis peapendicular tothe z-axis.

Answer»

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

so that it is perpendicular to axis Z.