If the vector \( \hat{\imath}-ax\,\hat{\jmath}+y\,\hat{k} \) and \( \hat{\imath}+x\,\hat{\jmath}-by\,\hat{k} \) are perpendicular, where \( a,b\in R \) and \( R \) is the set of real numbers then
Match List-I with List-II
| List-I | List-II | ||
| Condition on \( a \) and \( b \) | locus of \( (x,y) \) represents | ||
| (A) | \( a=b,\ a\gt 0,\ b\gt 0 \) | (I) | Ellipse |
| (B) | \( a\ne b,\ a\gt 0,\ b\gt 0 \) | (II) | A pair of straight lines |
| (C) | \( a\ne b,\ a.b\lt 0 \) | (III) | Circle |
| (D) | Either \( a=0 \) and \( b\gt 0 \) or \( b=0 \) and \( a\gt 0 \) | (IV) | Hyperbola |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A) - (III), (B) - (I), (C) - (IV), (D) - (II)
Perpendicular vectors have zero dot product:
\( \displaystyle 1\cdot 1+(-ax)(x)+(y)(-by)=0 \)
\( \displaystyle ax^{2}+by^{2}=1 \)
(A) \( a=b,\ a\gt 0,\ b\gt 0 \)
This matches (III). The equation becomes \( x^{2}+y^{2}=\frac{1}{a} \), a circle.
(B) \( a\ne b,\ a\gt 0,\ b\gt 0 \)
This matches (I). Both coefficients are positive and unequal, so \( ax^{2}+by^{2}=1 \) is an ellipse.
(C) \( a\ne b,\ a.b\lt 0 \)
This matches (IV). The coefficients have opposite signs, so \( ax^{2}+by^{2}=1 \) is a hyperbola.
(D) Either \( a=0 \) and \( b\gt 0 \) or \( b=0 \) and \( a\gt 0 \)
This matches (II). With a = 0 the equation gives \( y=\pm\frac{1}{\sqrt{b}} \); with b = 0 it gives \( x=\pm\frac{1}{\sqrt{a}} \). Either way it is a pair of straight lines.
So the matching is (A) - (III), (B) - (I), (C) - (IV), (D) - (II).