JEE Main 2017 · 8 Apr · Shift 1 · Mathematics Q83

← JEE Main 2017 · 8 Apr · Shift 1

MathematicsSingle correct+4 / -1Question 83 of 90

The line of intersection of the planes $\vec r\cdot(3\hat i-\hat j+\hat k)=1$ and $\vec r\cdot(\hat i+4\hat j-2\hat k)=2$ , is :

\dfrac{x-\frac47}{-2}=\dfrac y7=\dfrac{z-\frac57}{13}

\dfrac{x-\frac47}2=\dfrac y{-7}=\dfrac{z+\frac57}{13}

\dfrac{x-\frac6{13}}2=\dfrac{y-\frac5{13}}{-7}=\dfrac z{-13}

\dfrac{x-\frac6{13}}2=\dfrac{y-\frac5{13}}7=\dfrac z{-13}

Answer and worked solution

Official answer: C

The direction is $(3,-1,1)\times(1,4,-2)=(-2,7,13)$ . The point $\left(\frac6{13},\frac5{13},0\right)$ lies on both planes ( $\frac{18-5}{13}=1$ and $\frac{6+20}{13}=2$ ), so (3) is the line. The point in (1) and (2), $\left(\frac47,0,\pm\frac57\right)$ , is not on the first plane.

Topics:Equation of a line in space Cross product

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JEE Main 2017 · 8 Apr · Shift 1, Mathematics question 83