JEE Mathematics Three Dimensional Geometry MCQs Set 27

Answer: (c) are collinear

Explanation:
1. The vectors from the first point are \( (-3,-5,-3) \) and \( (3,5,3) \).
2. They are negatives of each other, so the three points lie on one line.

Teacher's Note:
1. Check whether the two displacement vectors are parallel.

Answer: (b) \( \frac{-2}{\sqrt{17}},\frac{-2}{\sqrt{17}},\frac3{\sqrt{17}} \)

Explanation:
1. \( \overrightarrow{AB}=(-4,-4,6) \) with length \( \sqrt{16+16+36}=2\sqrt{17} \).
2. Dividing: \( \left(\frac{-2}{\sqrt{17}},\frac{-2}{\sqrt{17}},\frac{3}{\sqrt{17}}\right) \).

Teacher's Note:
1. Divide the direction ratios by the length.

Answer: (c) 2

Explanation:
1. \( \cos^2\alpha+\cos^2\beta+\cos^2\gamma=1 \).
2. So \( \sin^2\alpha+\sin^2\beta+\sin^2\gamma=3-1=2 \).

Teacher's Note:
1. Use \( l^2+m^2+n^2=1 \).

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

Explanation:
1. \( m=n=\cos\frac\pi4=\frac1{\sqrt2} \), so \( m^2+n^2=1 \).
2. Then \( l^2=1-1=0 \), so the angle with the x-axis is \( \frac\pi2 \).

Teacher's Note:
1. Use \( l^2+m^2+n^2=1 \).

Answer: (a) \( (3,4,5) \)

Explanation:
1. In a parallelogram PQRS the diagonals PR and QS bisect each other.
2. So \( P+R=Q+S \) and \( S=P+R-Q=(4+1-2,\,5+2-3,\,10-1-4)=(3,4,5) \).

Teacher's Note:
1. Diagonals of a parallelogram bisect each other.

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

Explanation:
1. Direction ratios: \( (2-6,-3+7,1+1)=(-4,4,2) \), proportional to \( (-2,2,1) \) with length 3.
2. The direction cosines are \( \pm\left(-\frac23,\frac23,\frac13\right) \). For an acute angle with the x-axis, \( l>0 \), so we take \( \left(\frac23,-\frac23,-\frac13\right) \).

Teacher's Note:
1. An acute angle with the x-axis means a positive x-direction cosine.

Answer: (b) \( \frac43 \)

Explanation:
1. The diagonals have direction cosines \( \frac1{\sqrt3}(\pm1,\pm1,\pm1) \); for a line with direction cosines \( (l,m,n) \) the four cosines are \( \frac{\pm l\pm m\pm n}{\sqrt3} \).
2. Summing the squares: \( \frac{4(l^2+m^2+n^2)}3=\frac43 \).

Teacher's Note:
1. The cross terms cancel when the four squares are added.

Answer: (b) \( -1 \)

Explanation:
1. \( \cos2\theta=2\cos^2\theta-1 \), so the sum is \( 2(l^2+m^2+n^2)-3 \).
2. \( =2-3=-1 \).

Teacher's Note:
1. Use the double angle formula for cosine.

Answer: (a) 5

Explanation:
1. The projections are \( l\,d \), \( m\,d \), \( n\,d \), so the length satisfies \( d^2=3^2+1^2+15=25 \).
2. So \( d=5 \).

Teacher's Note:
1. Square and add the projections.

Answer: (d) \( (4,3,-2) \)

Explanation:
1. The normal is the cross product \( (1,-2,-1)\times(3,-2,3) \).
2. \( =(-6-2,\ -3-3,\ -2+6)=(-8,-6,4) \), proportional to \( (4,3,-2) \).

Teacher's Note:
1. Take the cross product of the two direction vectors.

Answer: (c) \( (6\sqrt2,6,\pm6) \)

Explanation:
1. \( l=\cos45^\circ=\frac1{\sqrt2} \), \( m=\cos60^\circ=\frac12 \), so \( n^2=1-\frac12-\frac14=\frac14 \) and \( n=\pm\frac12 \).
2. \( P=12(l,m,n)=(6\sqrt2,6,\pm6) \).

Teacher's Note:
1. Find the third direction cosine, then multiply by 12.

Answer: (a) \( l=m=n=\frac1{\sqrt3} \)

Explanation:
1. Equal angles give \( l=m=n \), so \( 3l^2=1 \).
2. In the first octant all direction cosines are positive, so \( l=m=n=\frac1{\sqrt3} \).

Teacher's Note:
1. The octant OXYZ is the first octant.

Answer: (a) \( \frac{\sqrt{23}}6 \)

Explanation:
1. \( \frac14+\frac19+n^2=1 \) gives \( n^2=1-\frac{13}{36}=\frac{23}{36} \).
2. So \( n=\frac{\sqrt{23}}{6} \) (taking the positive value).

Teacher's Note:
1. The squares of direction cosines add to 1.

Answer: (b) \( (-1,2,-2) \)

Explanation:
1. The length of the ratio vector is \( \sqrt{1+4+4}=3 \), so the direction cosines are \( \left(-\frac13,\frac23,-\frac23\right) \).
2. \( P=3\times \) these \( =(-1,2,-2) \).

Teacher's Note:
1. Multiply the direction cosines by OP.

Answer: (a) \( \frac\pi3 \)

Explanation:
1. \( lm=0 \) gives \( l=0 \) or \( m=0 \). If \( l=0 \): \( m=-n \) and the line is \( \left(0,\frac1{\sqrt2},-\frac1{\sqrt2}\right) \). If \( m=0 \): \( l=-n \) and the line is \( \left(\frac1{\sqrt2},0,-\frac1{\sqrt2}\right) \).
2. \( \cos\theta=0+0+\frac12=\frac12 \), so \( \theta=\frac\pi3 \).

Teacher's Note:
1. List the two possible lines and take their dot product.

Answer: (a) \( \frac3{\sqrt{70}}x+\frac5{\sqrt{70}}y-\frac6{\sqrt{70}}z=7 \)

Explanation:
1. The length of the normal vector is \( \sqrt{9+25+36}=\sqrt{70} \), so the unit normal is \( \frac{3\hat i+5\hat j-6\hat k}{\sqrt{70}} \).
2. The plane is \( \mathbf r\cdot\hat n=7 \), that is \( \frac{3x+5y-6z}{\sqrt{70}}=7 \).

Teacher's Note:
1. Use the form \( \mathbf r\cdot\hat n=d \) with a unit normal.

Answer: (c) \( \left(\frac{24}{29},\frac{36}{29},\frac{48}{29}\right) \)

Explanation:
1. The foot is \( t(2,3,4) \) with \( 2(2t)+3(3t)+4(4t)=12 \), so \( 29t=12 \) and \( t=\frac{12}{29} \).
2. The foot is \( \left(\frac{24}{29},\frac{36}{29},\frac{48}{29}\right) \).

Teacher's Note:
1. The foot lies along the normal direction from the origin.

Answer: (a) \( 2x+3y-3z=5 \)

Explanation:
1. The vectors in the plane are \( (0,1,1) \) and \( (-3,1,-1) \); their cross product is \( (-2,-3,3) \).
2. The plane is \( 2x+3y-3z=d \) and the point \( (1,1,0) \) gives \( d=5 \). Check with \( (1,2,1) \): \( 2+6-3=5 \).

Teacher's Note:
1. Find a normal by a cross product, then use one point.

Answer: (b) \( 7x-5y+4z-8=0 \)

Explanation:
1. The family is \( (3x-y+2z-4)+\lambda(x+y+z-2)=0 \). At \( (2,2,1) \): \( 2+3\lambda=0 \), so \( \lambda=-\frac23 \).
2. Then \( 3(3x-y+2z-4)-2(x+y+z-2)=7x-5y+4z-8=0 \).

Teacher's Note:
1. Use the family of planes through the line of intersection.

Answer: (a) \( \cos^{-1}\left(\frac{15}{\sqrt{731}}\right) \)

Explanation:
1. \( \mathbf n_1\cdot\mathbf n_2=6-6-15=-15 \), \( |\mathbf n_1|=\sqrt{17} \), \( |\mathbf n_2|=\sqrt{43} \).
2. \( |\cos\theta|=\frac{15}{\sqrt{731}} \), so the acute angle is \( \cos^{-1}\frac{15}{\sqrt{731}} \).

Teacher's Note:
1. The angle between planes is the angle between their normals.

Answer: (c) non-parallel, non-perpendicular and have an angle \( \cos^{-1}\left(\frac25\right) \)

Explanation:
1. \( \mathbf n_1\cdot\mathbf n_2=21-5-60=-44 \), and \( |\mathbf n_1|=|\mathbf n_2|=\sqrt{110} \).
2. \( \cos\theta=\frac{44}{110}=\frac25 \), so the planes are neither parallel nor perpendicular.

Teacher's Note:
1. Compare the normals \( (7,5,6) \) and \( (3,-1,-10) \).

Answer: (b) \( \frac{13}3 \) units

Explanation:
1. Distance \( =\frac{|2(3)+2+2(1)+3|}{\sqrt{4+1+4}}=\frac{13}{3} \).

Teacher's Note:
1. Use the point-to-plane distance formula.

Answer: (c) \( 51x+15y-50z+173=0 \)

Explanation:
1. The plane is \( (x+2y+3z-4)+\lambda(2x+y-z+5)=0 \) with normal \( (1+2\lambda,2+\lambda,3-\lambda) \).
2. Perpendicular to \( (5,3,6) \): \( 5+10\lambda+6+3\lambda+18-6\lambda=29+7\lambda=0 \), so \( \lambda=-\frac{29}7 \).
3. Substituting: \( 7(x+2y+3z-4)-29(2x+y-z+5)=0 \), i.e. \( -51x-15y+50z-173=0 \), or \( 51x+15y-50z+173=0 \).

Teacher's Note:
1. Use the dot product of normals to find \( \lambda \).

Answer: (b) \( \left(0,\frac{17}2,\frac{-13}2\right) \)

Explanation:
1. The line is \( (5,1,6)+t(-2,3,-5) \). The YZ-plane is \( x=0 \), so \( 5-2t=0 \) and \( t=\frac52 \).
2. Then \( y=1+\frac{15}{2}=\frac{17}2 \) and \( z=6-\frac{25}{2}=-\frac{13}2 \).

Teacher's Note:
1. The YZ-plane is where \( x=0 \).

Answer: (a) \( x+2y-3z-14=0 \)

Explanation:
1. The normal is \( \overrightarrow{OP}=(1,2,-3) \), so the plane is \( x+2y-3z=d \).
2. Through P: \( d=1+4+9=14 \).

Teacher's Note:
1. OP itself is the normal.

Answer: (d) \( \frac2{\sqrt{29}} \) units

Explanation:
1. Rewrite the second plane as \( 2x+3y+4z=6 \).
2. Distance \( =\frac{|6-4|}{\sqrt{4+9+16}}=\frac2{\sqrt{29}} \).

Teacher's Note:
1. Make the normal vectors identical first.

Answer: (b) parallel

Explanation:
1. The normals \( (2,-1,4) \) and \( (5,-2.5,10) \) are proportional (ratio 2.5).
2. The constants are not in the same ratio (\( 5\times2.5=12.5\ne6 \)), so the planes are parallel and distinct.

Teacher's Note:
1. Proportional normals mean parallel planes.

Answer: (b) \( r^2x+ry+z=3r^2 \)

Explanation:
1. If the intercepts are \( a,b,c \), the centroid is \( \left(\frac a3,\frac b3,\frac c3\right) \), so \( a=3 \), \( b=3r \), \( c=3r^2 \).
2. The plane \( \frac x3+\frac y{3r}+\frac z{3r^2}=1 \) becomes \( r^2x+ry+z=3r^2 \).

Teacher's Note:
1. Multiply the intercept form by \( 3r^2 \).