JEE Mathematics Three Dimensional Geometry MCQs Set 24

Answer: (b) \( 2 \)

Explanation:
1. A line is parallel to a plane iff its direction is perpendicular to the normal.
2. So \( (2,1,2)\cdot(3,-2,-m)=6-2-2m=0 \), giving \( m=2 \).

Teacher's Note:
1. Parallel means direction dot normal equals zero.
2. Solve the linear equation for \( m \).

Answer: (a) \( 2x-4y+3z-8=0 \)

Explanation:
1. Normal \( (1,2,2)\times(3,3,2)=(4-6,\ 6-2,\ 3-6)=(-2,4,-3) \), i.e. \( (2,-4,3) \).
2. Plane \( 2x-4y+3z=2+12-6=8 \), so \( 2x-4y+3z-8=0 \).

Teacher's Note:
1. A plane perpendicular to two planes has the cross product of their normals as normal.
2. Substitute the point to find the constant.

Answer: (c) (2) and (3) only

Explanation:
1. The cubic factorises: \( x^2(x+1)-4(x+1)=(x+1)(x-2)(x+2) \), so \( \ell=2,\ m=-1,\ n=-2 \).
2. Direction ratios \( (2,-1,-2) \) and \( (-1,-2,2) \): \( \cos\theta=\frac{|-2+2-4|}{9}=\frac49 \).
3. Then \( \sin\theta=\frac{\sqrt{65}}{9} \) (option 2) and \( \cos2\phi=2\cdot\frac{13}{18}-1=\frac49 \) for \( \phi=\cos^{-1}\sqrt{13/18} \) (option 3).

Teacher's Note:
1. Factorise the cubic by grouping.
2. The same angle can be written in several forms; test each option's cosine.

Answer: (c) (2) and (4) only

Explanation:
1. Plane: \( (2+2\lambda)-(-1+4\lambda)+(2+12\lambda)=5+10\lambda=5 \), so \( \lambda=0 \) and the point is \( (2,-1,2) \).
2. Distance squared: \( (\alpha-2)^2+(5\alpha+1)^2+(10\alpha-2)^2=126\alpha^2-34\alpha+9=169 \).
3. So \( 63\alpha^2-17\alpha-80=0 \), giving \( \alpha=-1 \) or \( \frac{80}{63} \).

Teacher's Note:
1. Find the intersection point first.
2. Then solve the resulting quadratic in \( \alpha \).

Answer: (a) (1) and (2) only

Explanation:
1. Use the edges \( \overrightarrow{AD},\overrightarrow{BD},\overrightarrow{CD} \) with determinant 0: \( xyz-xy-yz-zx=0 \).
2. Divide by \( xyz \): \( 1-\frac1z-\frac1x-\frac1y=0 \), so (1) is true.
3. (2) is equivalent: \( \sum\frac{x-1}{x}=3-\sum\frac1x=2 \). (3) and (4) do not follow.

Teacher's Note:
1. Set the scalar triple product of three edges to zero.
2. Divide by xyz to reach the reciprocal form.

Answer: (c) Either of the two:<br />rows \( (x-x_1,\ y-y_1,\ z-z_1),\ (x_2-x_1,\ y_2-y_1,\ z_2-z_1),\ (d_1,d_2,d_3) \) or rows \( (x-x_2,\ y-y_2,\ z-z_2),\ (x_1-x_2,\ y_1-y_2,\ z_1-z_2),\ (d_1,d_2,d_3) \)

Explanation:
1. The plane contains the point \( A \), the point \( (x_2,y_2,z_2) \) on the line and the direction \( (d_1,d_2,d_3) \).
2. A general point is in the plane iff the vectors from \( A \) (or from the point on the line) to it, to the other point, and the direction are coplanar.
3. Using either reference point gives the same plane, so both determinants are correct.

Teacher's Note:
1. Coplanarity of three vectors is a zero determinant.
2. Either given point on the plane can serve as the reference.

Answer: (d) \( \frac{x+1}{1}=\frac{y+1}{5}=\frac{z+1}{2} \)

Explanation:
1. Point of the line: \( (1+2t,2+3t,3+4t) \). On the plane: \( 7+7t=0 \), so \( t=-1 \) and \( A=(-1,-1,-1) \).
2. The least inclination line in the plane is the projection of the given line: \( (2,3,4)-\frac{7}{6}(1,-1,2)=\frac16(5,25,10) \), direction \( (1,5,2) \).
3. Line \( \frac{x+1}{1}=\frac{y+1}{5}=\frac{z+1}{2} \).

Teacher's Note:
1. The line of greatest slope or least inclination in a plane is the projection of the given line onto the plane.
2. Subtract the normal component of the direction.

Answer: (d) O and P both lie in the obtuse angle between the planes

Explanation:
1. Make both constants positive: the second plane becomes \( -x+2y-3z+5=0 \).
2. Dot product of normals \( 2(-1)+3(2)+(-4)(-3)=16>0 \), so \( O \) is in the obtuse angle.
3. At \( P \): \( 2-6-12+7=-9 \) and \( -1-4-9+5=-9 \); the product is positive, so \( P \) is on the same side as \( O \): both obtuse.

Teacher's Note:
1. First rewrite each plane so that the constant term is positive.
2. The sign of the dot product of normals decides where the origin lies.

Answer: (c) O lies in the acute angle and P in the obtuse angle

Explanation:
1. Both constants are positive. Dot product of normals \( 2+2-9=-5<0 \), so \( O \) lies in the acute angle.
2. At \( P \): \( 2-2-6+5=-1 \) and \( 4-1+6+1=10 \); the product is negative.
3. So \( P \) is on the opposite type: \( P \) lies in the obtuse angle.

Teacher's Note:
1. Check the sign of the constants first, then the dot product of the normals.
2. A negative product at \( P \) means \( P \) is not in the same region as \( O \).

Answer: (c) O and P both lie in the acute angle between the planes

Explanation:
1. Both constants are positive. Dot product of normals \( 1-4-9=-12<0 \), so \( O \) lies in the acute angle.
2. At \( P \): \( 1+4-6+2=1 \) and \( 1-4+6+7=10 \); the product is positive.
3. So \( P \) is on the same side as \( O \): both lie in the acute angle.

Teacher's Note:
1. Compute the dot product of normals and the product of the two plane values at \( P \).
2. A positive product means the same region as the origin.

Answer: (b) \( \frac{12}{\sqrt{10}} \)

Explanation:
1. Triangle ABD has a right angle at \( A \), so its orthocentre is \( A(0, 0, 0) \).
2. Plane EBC: \( \overrightarrow{BC}=(0,0,4) \), \( \overrightarrow{BE}=(-2,6,6) \), normal \( (-24,-8,0)\parallel(3,1,0) \); through \( B \): \( 3x+y=12 \).
3. Distance from the origin \( =\frac{12}{\sqrt{10}} \).

Teacher's Note:
1. The orthocentre of a right-angled triangle is at the right-angle vertex.
2. Find the plane equation from a normal and one point.

Answer: (a) \( x^2+y^2+z^2-x-2y-3z=0 \)

Explanation:
1. If \( F \) is the foot, then \( \overrightarrow{OF}\perp\overrightarrow{FP} \) where \( P = (1, 2, 3) \), so \( F \) lies on the sphere with diameter OP.
2. Equation \( x(x-1)+y(y-2)+z(z-3)=0 \), i.e. \( x^2+y^2+z^2-x-2y-3z=0 \).

Teacher's Note:
1. The angle at the foot of a perpendicular is a right angle.
2. A point seeing a segment at a right angle lies on the sphere with that segment as diameter.

Answer: (c) \( \sqrt{56} \)

Explanation:
1. Centre \( \left(1,1,\frac{11}{3}\right) \), radius squared \( 1+1+\frac{121}{9}=\frac{139}{9} \).
2. The point \( (1,2,3) \) is on the line; \( \overrightarrow{CP}=\left(0,1,-\frac23\right) \). Cross with direction \( (1,2,3) \): \( \left(\frac{13}{3},-\frac23,-1\right) \), squared length \( \frac{182}{9} \).
3. Distance squared from centre to line \( =\frac{182}{9\cdot14}=\frac{13}{9} \).
4. Chord \( =2\sqrt{\frac{139}{9}-\frac{13}{9}}=2\sqrt{14}=\sqrt{56} \).

Teacher's Note:
1. Chord = \( 2 \sqrt{r^2 - d^2} \) where \( d \) is the distance of the centre from the line.
2. Use the cross product formula for \( d \).

Answer: (b) \( \left(\frac83,\frac43,-\frac73\right) \)

Explanation:
1. Value at \( P \): \( 1-4-2=-5 \). Distance \( \frac{|-5-\alpha|}{3}=5 \), so \( \alpha=10 \) (positive).
2. Foot \( =P-\frac{(-5-10)}{9}(1,2,-2)=P+\frac53(1,2,-2) \).
3. So the foot is \( \left(\frac83,\frac43,-\frac73\right) \).

Teacher's Note:
1. Find \( \alpha \) from the distance, keeping \( \alpha \) positive.
2. Foot = \( P \) - (value / |n|^2) n.

Answer: (c) \( \frac1{\sqrt2} \)

Explanation:
1. Line QR: \( (2-t,\ 3-4t,\ 5-t) \). On the plane: \( 10-5t-12+16t-5+t=1 \), so \( 12t=8,\ t=\frac23 \); \( P \) at \( t=\frac23 \).
2. Foot of \( T \): \( \overrightarrow{QT}=(0,-2,-1) \), direction \( (-1,-4,-1) \): \( t=\frac{(0,-2,-1)\cdot(-1,-4,-1)}{18}=\frac{9}{18}=\frac12 \); \( S \) at \( t=\frac12 \).
3. PS \( =\left|\frac23-\frac12\right|\sqrt{18}=\frac16\cdot3\sqrt2=\frac1{\sqrt2} \).

Teacher's Note:
1. Both \( P \) and \( S \) are points on the same line, so find their parameters.
2. The distance is the difference of parameters times the length of the direction vector.

Answer: (a) (2) and (3) only

Explanation:
1. Points \( (1,-1,0) \) and \( (-1,-1,0) \); join \( (-2,0,0) \). Coplanarity: \( \begin{vmatrix}2&k&2\\5&2&k\\-2&0&0\end{vmatrix}=0 \) gives \( k^2=4 \), so \( k=\pm2 \).
2. For \( k=2 \): normal \( (2,2,2)\times(5,2,2)=(0,6,-6) \), plane \( y-z=-1 \).
3. For \( k=-2 \): normal \( (2,-2,2)\times(5,2,-2)=(0,14,14) \), plane \( y+z=-1 \).

Teacher's Note:
1. Coplanarity gives the possible values of \( k \) first.
2. For each value, the normal is the cross product of the two directions.

Answer: (d) (2) and (4) only

Explanation:
1. \( P3 \): \( x+z-1+ky=0 \). Distance of \( (0,1,0) \): \( \frac{|k-1|}{\sqrt{2+k^2}}=1 \), so \( k=-\frac12 \) and \( P3 \) is \( 2x-y+2z-2=0 \).
2. Distance of \( (\alpha,\beta,\gamma) \): \( \frac{|2\alpha-\beta+2\gamma-2|}{3}=2 \), so \( 2\alpha-\beta+2\gamma-2=\pm6 \).
3. These are \( 2\alpha-\beta+2\gamma-8=0 \) and \( 2\alpha-\beta+2\gamma+4=0 \).

Teacher's Note:
1. A plane through the intersection of two planes is a combination of them.
2. Use the distance formula twice.

Answer: (a) (1) and (2) only

Explanation:
1. A line at constant distance from both planes is parallel to both: direction \( (1,2,-1)\times(2,-1,1)=(1,-3,-5) \).
2. The feet of perpendiculars from \( L \) on \( P_1 \) form a line in \( P_1 \), parallel to \( L \), through the foot of the origin: \( -\frac{1}{6}(1,2,-1)=\left(-\frac16,-\frac13,\frac16\right) \).
3. Points of \( M \): \( \left(-\frac16,-\frac13,\frac16\right)+t(1,-3,-5) \). Check (1): \( t=\frac16 \) gives \( \left(0,-\frac56,-\frac23\right) \) - yes. (2) is the case \( t=0 \).

Teacher's Note:
1. A line equidistant from two planes through it must be parallel to both.
2. The locus of feet is the projection of \( L \) on the plane.

Answer: (c) \( (2,8,-3),\ (0,1,2),\ \sqrt{78} \)

Explanation:
1. Points: \( (5+3s,\ 7-s,\ -2+s) \) and \( (-3-3t,\ 3+2t,\ 6+4t) \). Their difference must be proportional to \( (2,7,-5) \).
2. Solving gives \( s=t=-1 \): the points are \( (2,8,-3) \) and \( (0,1,2) \).
3. Length \( =\sqrt{4+49+25}=\sqrt{78} \).

Teacher's Note:
1. Set the difference of the two general points equal to a multiple of the given direction.
2. Solve the three equations for \( s, t \) and the multiple.

Answer: (b) \( x-2y+z=0 \)

Explanation:
1. Points \( (a-d,a,a+d) \) and \( (b-c,b,b+c) \) lie on the plane \( x-2y+z=0 \) (since \( (a-d)-2a+(a+d)=0 \)).
2. Directions \( (\alpha-\delta,\alpha,\alpha+\delta) \) and \( (\beta-\gamma,\beta,\beta+\gamma) \) satisfy \( x-2y+z=0 \) too.
3. Both lines lie in this plane and, as they are not parallel, it is the plane they determine.

Teacher's Note:
1. Check whether a simple plane contains both points and both directions.
2. The coefficient pattern (1, -2, 1) makes every vector of the form \( (p-q, p, p+q) \) lie in the plane.

Answer: (d) \( 9(x^2+y^2+z^2)=5 \)

Explanation:
1. Write the line as \( \frac{x+1}{1/2}=\frac{y-2}{-1}=\frac{z+3}{1} \): direction \( (1,-2,2) \), point \( (-1,2,-3) \).
2. Distance from the origin \( =\frac{|(-1,2,-3)\times(1,-2,2)|}{3}=\frac{|(-2,-1,0)|}{3}=\frac{\sqrt5}{3} \).
3. Radius squared \( \frac59 \): \( 9(x^2+y^2+z^2)=5 \).

Teacher's Note:
1. Rewrite the line in standard symmetric form before reading off a direction.
2. The radius is the distance from the centre to the line.

Answer: (d) \( -\frac13,-\frac23,\frac23 \)

Explanation:
1. The incident ray travels from the source on the x-axis to the origin, direction \( (-1,0,0) \). The unit normal is \( \hat n=\frac{(1,-1,1)}{\sqrt3} \).
2. Reflected direction: \( \vec d-2(\vec d\cdot\hat n)\hat n=(-1,0,0)+\frac23(1,-1,1)=\left(-\frac13,-\frac23,\frac23\right) \).
3. This is already a unit vector, giving the direction cosines.

Teacher's Note:
1. Reflection formula: d' = d - 2 (d . n) n for a unit normal \( n \).
2. Check that the result has unit length.