JEE Advanced 2019 · Paper 1 · Mathematics Q18

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MathematicsNumerical+3Question 54 of 54

Three lines are given by

$\vec r = \lambda\hat i,\ \lambda\in\mathbb R;\quad\vec r = \mu(\hat i + \hat j),\ \mu\in\mathbb R\quad\text{and}\quad\vec r = v(\hat i + \hat j + \hat k),\ v\in\mathbb R.$

Let the lines cut the plane $x + y + z = 1$ at the points $A$ , $B$ and $C$ respectively. If the area of the triangle $ABC$ is $\Delta$ then the value of $(6\Delta)^2$ equals ______

Your answer

Answer and worked solution

Official answer: 0.75

The points. $A=(1,0,0)$ , $B=\left(\frac12,\frac12,0\right)$ and $C=\left(\frac13,\frac13,\frac13\right)$ .

Cross product.
$\overrightarrow{AB}\times\overrightarrow{AC}=\left(-\tfrac12,\tfrac12,0\right)\times\left(-\tfrac23,\tfrac13,\tfrac13\right)=\left(\tfrac16,\tfrac16,\tfrac16\right).$

Area. $\Delta=\frac12\cdot\frac{\sqrt3}6=\frac{\sqrt3}{12}$ , so $(6\Delta)^2=\frac34$ .

Answer: 0.75

Topics:Cross product Equation of a plane

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