87917 If |a+b|=a−b∣, then a and b are
(B) : Given, |a+b|=|a−b|Squaring both sides,⇒|a+b|2=|a−b|2⇒|a|2+|b|2+2a⋅b=|a|2+|b|2−2a⋅b⇒4a⋅b=0a⋅b=0hence, a is perpendicular to b.
87952 The image of the point with position vector i^+3k^ in the plane r.(i^+j^+k^)=1 is
(B) : Here, the point is (1,0,3) and equation of plane is x+y+z=1Let, α,β and γ be the image of plane.∴α−11=β−01=γ−31=−2(1+0+3−1)12+12+12α−11=β1=γ−31=−63α=−1,β=−2,γ=1Hence, the image is −i^−2j^+k^
87926 The measure of the angle between the line r→=(2,−3,1)+k(2,2,1);k∈:R and the plane 2x −2y+z+7=0 is
(A) : Given that,sinϕ=|a→⋅b→|a→||b→||a→=(2,2,1) and b→=(2,−2,1)=|(2,2,1)22+22+1222+(−2)2+12|=|4−4+14+4+14+4+1|=|19⋅9|sinϕ=19ϕ=sin−1(19)∴ Pythagoras theorem,(AB)2+(BC)2=AC2(BC)2=(9)2−(1)2(BC)2=81−1=80BC=45Then,tan−1(145)
87901 The value of x if x(i^+j^+k^) is a unit vector is
(A) : Given,A is unit vectors,x(i^+j^+k^)x(i^+j^+k^)=xi^+xj^+xk^|xi^+xj^+xk^|=1|(x)2+(x)2+(x)2|=1|3x2|=1⇒|x|3=1⇒|x|=13x=±13
87902 If p→=xa→+yb→+zc→ and p→=3i^+2j^+4k^,where a→=i^+j^,b→=j^+k^,c→=k^+i^, then value of x,y,z are respectively
(D) : Given,p→=xa→+yb→+zc→p→=3i^+2i^+4k^a→=i^+j^b→=j^+k^Accordifig t¯b^+q^ estion,p→=xa→+yb→+zc→3i^+2j^+4k^=x(i^+j^)+y(j^+k^)+z(k^+i^)3i^+2j^+4k^=(x+z)i^+(x+y)j^+(y+z)k^On comparing both sides, we get-∴x+z=3....(i)x+y=2And y+z=4On adding equation (i), (ii) and (iii) we get 2(x+y+z)=9∴x+y+z=92∴x=92−4 and y=92−3And, z=92−2x=12,y=32,z=52