Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

3D Geometry question

2016 · Shift 1 · Q20
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Advanced
  3. /Mathematics
  4. /3D Geometry
  5. /2016 · Shift 1 · Q20

3D Geometry question

2016 · Shift 1 · Q20

JEE AdvancedMathematics3D GeometryMultiple correct+4 / −2
Consider a pyramid OPQRSOPQRSOPQRS located in the first octant (x≥0,y≥0,z≥0)\left( {x \ge 0,y \ge 0,z \ge 0} \right)(x≥0,y≥0,z≥0) with OOO as origin, and OPOPOP and OROROR along the xxx-axis and the yyy-axis, respectively. The base OPQROPQROPQR of the pyramid is a square with OP=3.OP=3.OP=3. The point SSS is directly above the mid-point, TTT of diagonal OQOQOQ such that TS=3.TS=3.TS=3. Then
  1. A
    the acute angle between OQOQOQ and OSOSOS is π3{\pi \over 3}3π​
  2. B
    the equation of the plane containing the triangle OQSOQSOQS is x−y=0x-y=0x−y=0
  3. C
    the length of the perpendicular from PPP to the plane containing the triangle OQSOQSOQS is 32{3 \over {\sqrt 2 }}2​3​
  4. D
    the perpendicular distance from OOO to the straight line containing RSRSRS is 152\sqrt {{{15} \over 2}}215​​
View written solutionFree

Correct answer: B, C, D

  1. Assign coordinates

Since OOO is the origin and the square OPQROPQROPQR lies in the xyxyxy-plane with:

  • OPOPOP along the xxx-axis,
  • OROROR along the yyy-axis,
  • OP=3OP=3OP=3,

we take O=(0,0,0),P=(3,0,0),R=(0,3,0),Q=(3,3,0).O=(0,0,0),\quad P=(3,0,0),\quad R=(0,3,0),\quad Q=(3,3,0).O=(0,0,0),P=(3,0,0),R=(0,3,0),Q=(3,3,0).

The midpoint of diagonal OQOQOQ is T=(32,32,0).T=\left(\frac{3}{2},\frac{3}{2},0\right).T=(23​,23​,0).

Point SSS is directly above TTT and TS=3TS=3TS=3, so S=(32,32,3).S=\left(\frac{3}{2},\frac{3}{2},3\right).S=(23​,23​,3).

  1. Check option A: acute angle between OQOQOQ and OSOSOS

Vectors: OQ→=(3,3,0),OS→=(32,32,3).\overrightarrow{OQ}=(3,3,0), \qquad \overrightarrow{OS}=\left(\frac{3}{2},\frac{3}{2},3\right).OQ​=(3,3,0),OS=(23​,23​,3).

Their dot product is OQ→⋅OS→=3⋅32+3⋅32+0⋅3=9.\overrightarrow{OQ}\cdot \overrightarrow{OS}=3\cdot \frac{3}{2}+3\cdot \frac{3}{2}+0\cdot 3=9.OQ​⋅OS=3⋅23​+3⋅23​+0⋅3=9.

Magnitudes: ∣OQ→∣=32+32=32,|\overrightarrow{OQ}|=\sqrt{3^2+3^2}=3\sqrt{2},∣OQ​∣=32+32​=32​,

=\sqrt{\frac{9}{4}+\frac{9}{4}+9} =\sqrt{\frac{27}{2}}=\frac{3\sqrt{6}}{2}.$$ So $$\cos\theta=\frac{9}{(3\sqrt{2})(3\sqrt{6}/2)} =\frac{9}{9\sqrt{3}}\cdot 2?$$ Let us simplify carefully: $$ (3\sqrt2)\left(\frac{3\sqrt6}{2}\right)=\frac{9\sqrt{12}}{2}=\frac{18\sqrt3}{2}=9\sqrt3. $$ Hence $$\cos\theta=\frac{9}{9\sqrt3}=\frac{1}{\sqrt3}. $$ Therefore $$\theta=\cos^{-1}\left(\frac{1}{\sqrt3}\right)\ne \frac{\pi}{3}.$$ So **A is false**. --- 3. **Check option B: equation of plane containing triangle $OQS$** Points in the plane are $$O=(0,0,0),\quad Q=(3,3,0),\quad S=\left(\frac{3}{2},\frac{3}{2},3\right).$$ Notice that for all three points, $x=y$. Thus these points lie on the plane $$x-y=0.$$ Also, the vectors $$\overrightarrow{OQ}=(3,3,0), \qquad \overrightarrow{OS}=\left(\frac{3}{2},\frac{3}{2},3\right)$$ are not multiples of each other, so they indeed determine a unique plane. Hence the plane of triangle $OQS$ is $$x-y=0.$$ So **B is true**. --- 4. **Check option C: perpendicular distance from $P$ to plane $OQS$** Plane $OQS$ is $$x-y=0.$$ Distance of point $P=(3,0,0)$ from plane $x-y=0$ is $$d=\frac{|3-0|}{\sqrt{1^2+(-1)^2}}=\frac{3}{\sqrt2}.$$ So **C is true**. --- 5. **Check option D: perpendicular distance from $O$ to line $RS$** Coordinates: $$R=(0,3,0),\qquad S=\left(\frac{3}{2},\frac{3}{2},3\right).$$ Direction vector of line $RS$: $$\overrightarrow{RS}=S-R=\left(\frac{3}{2},-\frac{3}{2},3\right).$$ Distance from point $O$ to line through $R$ and $S$ is $$d=\frac{|\overrightarrow{OR}\times \overrightarrow{RS}|}{|\overrightarrow{RS}|},$$ where $$\overrightarrow{OR}=(0,3,0).$$ Compute cross product: $$\overrightarrow{OR}\times \overrightarrow{RS} =\begin{vmatrix} \hat i & \hat j & \hat k\\ 0 & 3 & 0\\ \frac32 & -\frac32 & 3 \end{vmatrix} =\left(9,0,-\frac{9}{2}\right).$$ Its magnitude is $$\sqrt{9^2+0^2+\left(-\frac92\right)^2} =\sqrt{81+\frac{81}{4}} =\sqrt{\frac{405}{4}} =\frac{9\sqrt5}{2}.$$ Also, $$|\overrightarrow{RS}|=\sqrt{\left(\frac32\right)^2+\left(-\frac32\right)^2+3^2} =\sqrt{\frac94+\frac94+9} =\sqrt{\frac{27}{2}} =\frac{3\sqrt6}{2}.$$ Therefore $$d=\frac{\frac{9\sqrt5}{2}}{\frac{3\sqrt6}{2}}=\frac{3\sqrt5}{\sqrt6}=\frac{\sqrt{90}}{2?}$$ Simplify properly: $$d=\frac{3\sqrt5}{\sqrt6}=\frac{3\sqrt{30}}{6}=\frac{\sqrt{30}}{2}=\sqrt{\frac{15}{2}}.$$ So **D is true**. --- 6. **Final conclusion** - A: False - B: True - C: True - D: True Hence the correct options are $$\boxed{B,\ C,\ D}.$$ This matches the stored correct answer.
PreviousNext

More from 3D Geometry

  • Let P be the image of the point (3,1,7) with respect to the plane x−y+z=3. Then the equation of the plane passing through P and containing the straight line 1x​=2y​=1z​ is2016 · MCQ
  • In R3, consider the planes P1​:y=0 and P2​:x+z=1. Let P3​ be the plane, different from P1​ and P2​, which passes through the intersection of P1​ and P2​. If the distance of the point (0,1,0) from…2015 · Multiple correct
  • In R3, let L be a straight lines passing through the origin. Suppose that all the points on L are at a constant distance from the two planes P1​:x+2y−z+1=0 and P2​:2x−y+z−1=0. Let M be the locus of the…2015 · Multiple correct
  • From a point P(λ,λ,λ), perpendicular PQ and PR are drawn respectively on the lines y=x,z=1 and y=−x,z=−1. If P is such that ∠QPR is a right angle, then the possible value(s) of λ…2014 · Multiple correct
  • Perpendiculars are drawn from points on the line 2x+2​=−1y+1​=3z​ to the plane x+y+z=3. The foot of perpendiculars lie on the line2013 · MCQ
  • A line l passing through the origin is perpendicular to the lines l1​:(3+t)i+(−1+2t)j​+(4+2t)k,−∞<t<∞l2​:(3+2s)i+(3+2s)j​+(2+s)k,−∞<s<∞…2013 · Multiple correct
  • Consider the lines L1​:2x−1​=−1y​=1z+3​,L2​:1x−4​=1y+3​=2z+3​ and the planes P1​:7x+y+2z=3,P2​=3x+5y−6z=4. Let ax+by+cz=d be…2013 · MCQ
  • Two lines L1​:x=5,3−αy​=−2z​ and L2​:x=α,−1y​=2−αz​ are coplanar. Then α can take value(s)2013 · Multiple correct