3D Geometry · JEE Main previous year questions

2017 5 questions

1. Q232017 · 9 Apr 2017 · Shift 1

   If $x=a,y=b,z=c$ is a solution of the system of linear equations $x+8y+7z=0$ $9x+2y+3z=0$ $x+y+z=0$ such that the point (a, b, c) lies on the plane $x+2y+z=6$, then $2a+b+c$ equals :

   1. A.$-1$
   2. B.$0$
   3. C.$1$
   4. D.$2$
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   Answer: C

2. Q282017 · 9 Apr 2017 · Shift 1

   If a variable plane, at a distance of 3 units from the origin, intersects the coordinate axes at A, B and C, then the locus of the centroid of $\Delta \mathrm{ABC}$ is :

   1. A.$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=1$
   2. B.$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=3$
   3. C.$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{1}{9}$
   4. D.$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=9$
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   Answer: A

3. Q292017 · 9 Apr 2017 · Shift 1

   If the line, $\frac{x-3}{1}=\frac{y+2}{-1}=\frac{z+\lambda }{-2}$ lies in the plane, $2x-4y+3z=2$, then the shortest distance between this line and the line, $\frac{x-1}{12}=\frac{y}{9}=\frac{z}{4}$ is :

   1. A.$2$
   2. B.$1$
   3. C.$0$
   4. D.$3$
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   Answer: C

4. Q282017 · 8 Apr 2017 · Shift 1

   The coordinates of the foot of the perpendicular from the point $(1,-2,1)$ on the plane containing the lines, $\frac{x+1}{6}=\frac{y-1}{7}=\frac{z-3}{8}$ and $\frac{x-1}{3}=\frac{y-2}{5}=\frac{z-3}{7}$, is :

   1. A.$(2,-4,2)$
   2. B.$(-1,2,-1)$
   3. C.$(0,0,0)$
   4. D.$(1,1,1)$
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   Answer: C

5. Q292017 · 8 Apr 2017 · Shift 1

   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 :

   1. A.$\frac{x-\frac{4}{7}}{-2}=\frac{y}{7}=\frac{z-\frac{5}{7}}{13}$
   2. B.$\frac{x-\frac{4}{7}}{2}=\frac{y}{-7}=\frac{z+\frac{5}{7}}{13}$
   3. C.$\frac{x-\frac{6}{13}}{2}=\frac{y-\frac{5}{13}}{-7}=\frac{z}{-13}$
   4. D.$\frac{x-\frac{6}{13}}{2}=\frac{y-\frac{5}{13}}{7}=\frac{z}{-13}$
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   Answer: C