Find the vector equation of the line passing through the point (1, 2, -4) and perpendicular to the two lines
x−83=y+19−16=z−107 and x−153=y−298=z−5−5
Find the vector equation of the line passing through the point (1, 2, -4) and perpendicular to the two lines
x−83=y+19−16=z−107 and x−153=y−298=z−5−5
Any line through (1, 2, -4) can be written as
x−1a=y−2b=z+4c ...(i)
where, a, b, c are the direction ratios of line (i).
Now, the line (i) be perpendicular to the lines
x−83=y+19−16=z−107 and x−153=y−298=z−5−5
In above having direction ratios are (3, -16, 7) and (3, 8, -5) respectively.
Which is perpendicular with the Eq. (i).
∴ 3a - 16b + 7c = 0 ...(ii)
and 3a + 8b - 5c = 0 ...(iii)
By cross-multiplication, we have
a80−56=b21+15=c24+48⇒a24=b36=c72⇒a2=b3=c6=λ (say)∴ a=2λ, b=3λ and c=3λ
The equation of required line which is passes through the point (1, 2, -4) and parallel to vector 2^i+3^j+6^k is r=(^i+2^j−4^k)+λ(2^i+3^j+6^k)
Any line through (1, 2, -4) can be written as
x−1a=y−2b=z+4c ...(i)
where, a, b, c are the direction ratios of line (i).
Now, the line (i) be perpendicular to the lines
x−83=y+19−16=z−107 and x−153=y−298=z−5−5
In above having direction ratios are (3, -16, 7) and (3, 8, -5) respectively.
Which is perpendicular with the Eq. (i).
∴ 3a - 16b + 7c = 0 ...(ii)
and 3a + 8b - 5c = 0 ...(iii)
By cross-multiplication, we have
a80−56=b21+15=c24+48⇒a24=b36=c72⇒a2=b3=c6=λ (say)∴ a=2λ, b=3λ and c=3λ
The equation of required line which is passes through the point (1, 2, -4) and parallel to vector 2^i+3^j+6^k is r=(^i+2^j−4^k)+λ(2^i+3^j+6^k)
