A straight line cuts off the intercepts $OA = a$ and $OB = b$ on the positive directions of $x$-axis and $y$ axis respectively If the perpendicular from origin $O$ to this line makes an angle of $\frac{\pi}{6}$ with positive direction of $y$-axis and the area of $\triangle O A B$ is $\frac{98}{3} \sqrt{3}$, then $a^2-b^2$ is equal to :
- $\frac{392}{3}$
- $\frac{196}{3}$
- 196
- 98
The Correct Option is A
Solution and Explanation
Or
Comparing both :
Now area of
Top Questions on Coplanarity of Two Lines
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Concepts Used:
Coplanarity of Two Lines
Condition for Coplanarity Using Cartesian Form
The condition for coplanarity in the Cartesian form appears from the vector form.
Let's consider two points L (a1, b1, c1) & Q (a2, b2, c2) in the Cartesian plane,
Presuppose that there are two vectors q1 and q2. Their direction ratios are subjected by {x1, y1, z1}, and {x2, y2, z2} respectively.
The vector form of equation of the line in connection to L and Q can be stated as under:
LQ = (a2 – a1)i + (b2 – b1)j + (c2 – c1)k
Q1 = x1i + y1j + z1k
Q2 = x2i + y2j + z2k
Condition for Coplanarity Using Vector Form
For the derivation of the condition for coplanarity in vector form, we shall take into consideration the equations of two straight lines to be as stated below:
r1 = l1 + λq1
r2 = l2 + λq2
The condition for coplanarity in vector form is that the line in connection to the two points should be perpendicular to the product of the two vectors, q1 and q2.
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