JEE Mathematics Three Dimensional Geometry MCQs Set 05

Answer: (b) (1, 1, 1)

Explanation:
1. The intercepts are \( \frac1a,\frac1b,\frac1c \), so the sum of their reciprocals is \( a+b+c=1 \).
2. The point \( (1,1,1) \) satisfies \( ax+by+cz=a+b+c=1 \).

Teacher's Note:
1. Reciprocals of the intercepts are the coefficients a, b, c.
2. Check which point makes the plane equation hold for all such a, b, c.

Answer: (a) \( 2\sqrt3 \)

Explanation:
1. Distance \( =\frac{1}{\sqrt{4+9+\lambda^2}}=\frac15 \).
2. \( 13+\lambda^2=25\Rightarrow\lambda^2=12 \), so \( \lambda=2\sqrt3 \) since \( \lambda>0 \).

Teacher's Note:
1. Use the distance of a plane from the origin.
2. Square both sides and pick the positive root.

Answer: (c) 2x - 21y - 13z = 0

Explanation:
1. The plane passes through O, A and B, so its normal is \( \vec{OA}\times\vec{OB} \).
2. \( (1,-3,5)\times(4,1,-1)=(3-5,\ 20+1,\ 1+12)=(-2,21,13) \).
3. The plane is \( -2x+21y+13z=0 \), i.e. \( 2x-21y-13z=0 \).

Teacher's Note:
1. Three points including the origin fix the plane.
2. The cross product of two position vectors gives the normal.

Answer: (b) \( -\frac75 \)

Explanation:
1. The line lies in the plane if its direction is perpendicular to the normal: \( 3(1)-2(-2)+5\lambda=0\Rightarrow\lambda=-\frac75 \).
2. The point \( (1,-1,-1) \) lies on the plane: \( 3+2-5=0 \). So the condition is satisfied.

Teacher's Note:
1. Check both the direction and a point of the line against the plane.
2. Perpendicularity to the normal gives lambda.

Answer: (a) \( \frac{x-5}7=\frac{y-2}{-2}=\frac{z-4}3 \)

Explanation:
1. The perpendicular has direction \( (7,-2,3) \), the normal of the plane.
2. The line through \( (-2,4,1) \) is \( \frac{x+2}7=\frac{y-4}{-2}=\frac{z-1}3 \).
3. The point \( (5,2,4) \) gives \( \frac{7}{7}=\frac{-2}{-2}=\frac33=1 \), so it is on this line; option (a) is the same line.

Teacher's Note:
1. The perpendicular to a plane is parallel to its normal.
2. Different forms of the same line can use different points.

Answer: (b) \( \alpha^2+\beta^2+\gamma^2=\alpha'^2+\beta'^2+\gamma'^2 \)

Explanation:
1. The plane passes through the origin and is the perpendicular bisector of PQ.
2. Every point of the plane, including O, is equidistant from P and Q, so \( OP=OQ \).
3. Hence \( \alpha^2+\beta^2+\gamma^2=\alpha'^2+\beta'^2+\gamma'^2 \).

Teacher's Note:
1. A plane through the origin is the perpendicular bisector of a point and its image.
2. The origin is equidistant from both.

Answer: (a) x - y + 2z = 8

Explanation:
1. The plane is the perpendicular bisector of the segment joining the points.
2. Normal \( =(3-1,-1-1,5-1)=(2,-2,4)\propto(1,-1,2) \); midpoint \( =(2,0,3) \).
3. Plane: \( x-y+2z=2-0+6=8 \).

Teacher's Note:
1. The normal is along the line joining a point and its image.
2. The midpoint lies on the plane.

Answer: (b) x - y - 3z = 0

Explanation:
1. The planes through the intersection: \( (5x+y-3z-2)+\lambda(x+2y+3z-1)=0 \).
2. It passes through the origin: \( -2-\lambda=0\Rightarrow\lambda=-2 \).
3. Then \( 3x-3y-9z=0\Rightarrow x-y-3z=0 \).

Teacher's Note:
1. Use the family of planes through the line of intersection.
2. The origin fixes the parameter.

Answer: (c) 7(x - z) = 6

Explanation:
1. The family is \( (3x+y-5z-2)+\lambda(x-2y+3z-2)=0 \), normal \( (3+\lambda,\ 1-2\lambda,\ -5+3\lambda) \).
2. Perpendicular to the given plane: \( (3+\lambda)-(1-2\lambda)+(-5+3\lambda)=0\Rightarrow6\lambda=3\Rightarrow\lambda=\frac12 \).
3. The plane is \( \frac72x-\frac72z-3=0\Rightarrow7(x-z)=6 \).

Teacher's Note:
1. Two planes are perpendicular when their normals have zero dot product.
2. Use the family of planes through the line.

Answer: (d) \( \cos^{-1}\left(\frac25\sqrt{\frac23}\right) \)

Explanation:
1. \( \cos\theta=\frac{|3-4+5|}{\sqrt3\sqrt{50}}=\frac4{5\sqrt6}=\frac25\sqrt{\frac23} \).

Teacher's Note:
1. The angle between planes equals the angle between their normals.
2. Simplify 4 over 5 root 6.

Answer: (c) \( \frac x1=\frac y2=\frac{z-4}{-7} \)

Explanation:
1. Write it as \( k(2x-y)+(x+3y+z-4)=0 \). It always contains the line \( 2x-y=0,\ x+3y+z=4 \).
2. With \( y=2x \): \( 7x+z=4 \). Direction \( (1,2,-7) \) through \( (0,0,4) \).
3. The line is \( \frac x1=\frac y2=\frac{z-4}{-7} \).

Teacher's Note:
1. Collect the terms in the parameter k.
2. Solve the two fixed equations together.

Answer: (a) 11x - 3y = 0

Explanation:
1. A plane through the origin equally inclined to both has normal along \( \hat n_1\pm\hat n_2 \), where \( \hat n_1=\frac{(4,3,-5)}{5\sqrt2} \) and \( \hat n_2=\frac{(5,-12,13)}{13\sqrt2} \).
2. \( \hat n_1+\hat n_2\propto(52+25,\ 39-60,\ -65+65)=(77,-21,0)\propto(11,-3,0) \).
3. So \( 11x-3y=0 \) is equally inclined.

Teacher's Note:
1. Equally inclined planes have normals that bisect the given normals.
2. Use unit normals before adding.

Answer: (d) \( \frac2{\sqrt6},\frac{-1}{\sqrt6},\frac1{\sqrt6} \)

Explanation:
1. Direction \( (-2,1,-1) \) and normal \( (2,1,-3) \): dot product \( -4+1+3=0 \), so the line is parallel to the plane.
2. Its projection has the same direction, with magnitude \( \sqrt6 \).
3. Direction cosines: \( \pm\left(\frac2{\sqrt6},\frac{-1}{\sqrt6},\frac1{\sqrt6}\right) \), and option (d) is one of these. Option (c) is not proportional to \( (2,-1,1) \).

Teacher's Note:
1. A line parallel to a plane projects onto a parallel line.
2. Direction cosines can be given with either sign.

Answer: (c) aa' + cc' + 1 = 0

Explanation:
1. Taking y as the parameter, the direction ratios are \( (a,1,c) \) and \( (a',1,c') \).
2. Perpendicular means \( aa'+1+cc'=0 \).

Teacher's Note:
1. Write each line with y as the parameter.
2. Use zero dot product of directions.

Answer: (c) 2x - 10y + 5z = 51

Explanation:
1. On the line \( (1+2t,-t,-2+2t) \): the plane gives \( -1+3t=1\Rightarrow t=\frac23 \), so \( P=\left(\frac73,-\frac23,-\frac23\right) \).
2. The normal is \( \vec{PF}=\left(\frac23,-\frac{10}3,\frac53\right)\propto(2,-10,5) \) and the plane passes through \( (3,-4,1) \).
3. \( 2x-10y+5z=6+40+5=51 \).

Teacher's Note:
1. The foot of the perpendicular and P give the normal direction.
2. The plane passes through the foot.

Answer: (a) x + y + z = 0

Explanation:
1. The point \( (1,-1,0) \) and direction \( (1,-2,1) \) must satisfy the plane.
2. (a): \( 1-1+0=0 \) and \( 1-2+1=0 \), so it contains the line.
3. (b) gives \( 2-1=1\ne3 \); (c) gives \( 4-1=3\ne5 \); (d) gives \( 3-2=1\ne0 \). So only (a).

Teacher's Note:
1. A plane contains a line if it contains a point of the line and is parallel to its direction.
2. Test the point and the dot product with the normal.

Answer: (a) \( \frac49\sqrt{29} \)

Explanation:
1. Moving along direction \( (2,3,4) \): the point becomes \( (1+2t,\,2+3t,\,3+4t) \).
2. On the plane: \( 6+9t=2 \Rightarrow t=-\frac49 \).
3. Distance \( =|t|\sqrt{4+9+16}=\frac49\sqrt{29} \).

Teacher's Note:
1. Parametrise the line through the point with the given direction.
2. Distance equals the magnitude of the parameter times the length of the direction vector.

Answer: (a) \( \frac\pi2 \)

Explanation:
1. First line: \( \frac x{1/2}=\frac y{1/3}=\frac z{-1} \), direction \( (3,2,-6) \).
2. Second line: \( \frac x{1/6}=\frac y{-1}=\frac z{-1/4} \), direction \( (2,-12,-3) \).
3. Dot product \( =6-24+18=0 \), so the lines are perpendicular.

Teacher's Note:
1. Rewrite each line in symmetric form with unit coefficients.
2. Zero dot product means a right angle.

Answer: (a) \( \frac{10}{3\sqrt3} \)

Explanation:
1. Direction \( \cdot \) normal \( =1-5+4=0 \), so the line is parallel to the plane.
2. Distance of the point \( (2,-2,3) \) from the plane \( x+5y+z=5 \): \( \frac{|2-10+3-5|}{\sqrt{27}}=\frac{10}{3\sqrt3} \).

Teacher's Note:
1. A zero dot product of direction and normal means the line is parallel to the plane.
2. Then use the point to plane distance.

Answer: (b) \( 3a \)

Explanation:
1. Points: \( P=(t,t-a,t) \) on the first line and \( Q=(s-a,\frac s2,\frac s2) \) on the second.
2. \( P-Q=\left(t-s+a,\ t-a-\frac s2,\ t-\frac s2\right)\propto(2,1,2) \). Equating the ratios gives \( t=3a,\ s=2a \).
3. \( P-Q=(2a,a,2a) \) with length \( 3a \).

Teacher's Note:
1. Parametrise both lines and use the direction ratios of PQ.
2. Solve for the two parameters.

Answer: (a) (0, 1, 2)

Explanation:
1. PQ is along the normal \( (1,1,1) \), so \( Q=P-\lambda(1,1,1) \) with \( (1-\lambda)+(2-\lambda)+(3-\lambda)=3 \), giving \( \lambda=1 \) and \( Q=(0,1,2) \).
2. Check: \( 0+1+2=3 \) so Q lies on the plane; \( PQ=\sqrt3 \).

Teacher's Note:
1. Move from P along the normal by the signed distance.
2. Verify the point satisfies the plane equation.