A non-zero vector →a is parallel to the line of intersection of plane p1 determined by ^i+^j and ^i and plane P2 determined by vectors ^i−^j and ^i+^k, then angle between a and vector ^i−2^j+2^k is
Equation of the plane containing ^i and ^i+^j is
[(→r−^i)^i(^i+^j)]=0
⇒(→r−^i).[^i×(^i+^j)]=0
⇒{(x^i+y^j+z^k)−^i}.[^i×^i+^i×^j]=0
⇒{(x−1)^i+y^j+z^k}.[^k]=0⇒(x−1)^i.^k+y^j.^k+z^k.^k=0
⇒z=0 ...(i)
Equation of the plane containing ^i−^j and ^i+^k is
[→(r−(^i−^j))(^i−^j)(^i+^k)]=0
⇒(→r−^i+^j).[(^i−^j)×(^i+^k)]=0
⇒{(x^i+y^j+z^k)−(^i−^j)}.[^i×^i+^i×^k−^j×^i−^j×^k]=0
⇒{(x−1)^i+(y+1)^j+z^k}.[−^j+^k−^i]=0
⇒−(x−1)−(y+1)+z=0...(ii)
Let →a=a1^i+a2^j+a3^k
Since, →a is parallel to Eqs. (i) and (ii), we obtain
a3=0
And a1+a2−a3=0 ⇒a1=−a2,a3=0
Thus, vector in the direction →a is ^i−^j
If θ is the angle between →a and ^i−2^j+2^k, then
cosθ=±(1)(1)+(−1)(−2)√1+1√1+4+4=±3√2.3
⇒cosθ=±1√2⇒θ=π4or3π4