−−→AB=3^i−^j+^k and −−→CD=−3^i+2^j+4^k are two vectors.The position vectors of the points A and C are 6^i+7^j+4^k and −9^i+2^j,respectively. Find the position of vector of a point P on the line AB and a point Q on the line CD such that −−→PQ is perpendicular to −−→AB and −−→CD both.
−−→AB=3^i−^j+^k and −−→CD=−3^i+2^j+4^k are two vectors.The position vectors of the points A and C are 6^i+7^j+4^k and −9^i+2^j,respectively. Find the position of vector of a point P on the line AB and a point Q on the line CD such that −−→PQ is perpendicular to −−→AB and −−→CD both.
We have, −−→AB=3^i−^j+^k and −−→CD=−3^i+2^j+4^k
Also,the position vectors of A and C are 6^i+7^j+4^k and −9^i+2^j,respectively. Since,−−→PQ is perpendicular to both −−→AB and −−→CD.
So, P and Q will be the foot of perpendicular to both the lines through A and C.
Now,equation of the line through A and parallel to the vector −−→AB is,→r=(6^i+7^j+4^k)+λ(3^i−^j+^k)
and the line through C and parallel to the vector −−→CD is given by
→r=−9^i+2^j+μ(−3^i+2^j+4^k) .....(i)
Let →r=(6^i+7^j+4^k)+λ(3^i−^j+^k)
and →r=−9^i+2^j+μ(−3^i+2^j+4^k) ....(ii)
Let P(6+3λ,7−λ,4+λ) is any point on the first line and Q be any point on the second line is given by (−3μ,−9+2μ,2+4μ).
∴ −−→PQ=(−3μ−6−3λ)^i+(−9+2μ−7+λ)^j+(2+4μ−4−λ)^k=(−3μ−6−3λ)^i+(−2μ+λ−16)^j+(4μ−λ−2)^k
if −−→PQ is perpendicular to the first line,then
3(−3μ−6−3λ)−(2μ+λ−16)+(4μ−λ−2)=0⇒ −9μ−18−9λ−2μ−λ−32+16+4μ−λ−2=0⇒ −7μ−11λ−4=0 ...(iii)
if −−→PQ is perpendicular to the second line,then
3(−3μ−6−3λ)+2(2μ+λ−16)+4(4μ−λ−2)=0⇒ 9μ+18+9λ+4μ+2λ−32+16μ−4λ−8=0⇒ 29μ+7λ−22=0 ...(iv)
On solving Eqs.(iii) and (iv),we get
−49μ−77λ−28=0⇒ 319μ+77λ−242=0⇒ 270μ−270=0⇒ μ=1
Using μ in Eq.(iii),we get
−7(1)−11λ−4=0⇒−7−11λ−4=0⇒−11−11λ=0⇒ λ=−1∴−−→PQ=[−3(1)−6−3(−1)]^i+[2(1)+(−1)−16]^j+[4(1)−(−1)−2]^k=−6^i−15^j+3^k
We have, −−→AB=3^i−^j+^k and −−→CD=−3^i+2^j+4^k
Also,the position vectors of A and C are 6^i+7^j+4^k and −9^i+2^j,respectively. Since,−−→PQ is perpendicular to both −−→AB and −−→CD.
So, P and Q will be the foot of perpendicular to both the lines through A and C.
Now,equation of the line through A and parallel to the vector −−→AB is,→r=(6^i+7^j+4^k)+λ(3^i−^j+^k)
and the line through C and parallel to the vector −−→CD is given by
→r=−9^i+2^j+μ(−3^i+2^j+4^k) .....(i)
Let →r=(6^i+7^j+4^k)+λ(3^i−^j+^k)
and →r=−9^i+2^j+μ(−3^i+2^j+4^k) ....(ii)
Let P(6+3λ,7−λ,4+λ) is any point on the first line and Q be any point on the second line is given by (−3μ,−9+2μ,2+4μ).
∴ −−→PQ=(−3μ−6−3λ)^i+(−9+2μ−7+λ)^j+(2+4μ−4−λ)^k=(−3μ−6−3λ)^i+(−2μ+λ−16)^j+(4μ−λ−2)^k
if −−→PQ is perpendicular to the first line,then
3(−3μ−6−3λ)−(2μ+λ−16)+(4μ−λ−2)=0⇒ −9μ−18−9λ−2μ−λ−32+16+4μ−λ−2=0⇒ −7μ−11λ−4=0 ...(iii)
if −−→PQ is perpendicular to the second line,then
3(−3μ−6−3λ)+2(2μ+λ−16)+4(4μ−λ−2)=0⇒ 9μ+18+9λ+4μ+2λ−32+16μ−4λ−8=0⇒ 29μ+7λ−22=0 ...(iv)
On solving Eqs.(iii) and (iv),we get
−49μ−77λ−28=0⇒ 319μ+77λ−242=0⇒ 270μ−270=0⇒ μ=1
Using μ in Eq.(iii),we get
−7(1)−11λ−4=0⇒−7−11λ−4=0⇒−11−11λ=0⇒ λ=−1∴−−→PQ=[−3(1)−6−3(−1)]^i+[2(1)+(−1)−16]^j+[4(1)−(−1)−2]^k=−6^i−15^j+3^k
