JEE Mathematics Three Dimensional Geometry MCQs Set 21

Answer: (c) \( \frac\pi2 \)

Explanation:
1. For direction angles \( \alpha,\beta,\gamma \): \( \cos^2\alpha+\cos^2\beta+\cos^2\gamma=1 \).
2. With \( \alpha=\beta=\frac\pi4 \): \( \frac12+\frac12+\cos^2\gamma=1 \).
3. So \( \cos\gamma=0 \) and \( \gamma=\frac\pi2 \).

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

Answer: (a) \( 2\sqrt3\left(\hat i+\hat j+\hat k\right) \)

Explanation:
1. Equal inclination in the positive octant means all three direction cosines equal \( \frac1{\sqrt3} \).
2. So \( \vec r=6\cdot\frac1{\sqrt3}\left(\hat i+\hat j+\hat k\right) \).
3. \( \frac6{\sqrt3}=2\sqrt3 \), so \( \vec r=2\sqrt3\left(\hat i+\hat j+\hat k\right) \).

Teacher's Note:
1. Equally inclined and in the positive octant means the three components are equal and positive.

Answer: (a) \( -\frac{10}7 \)

Explanation:
1. The direction ratios are \( (-3,2k,2) \) and \( (3k,1,-5) \).
2. Perpendicular lines have zero dot product: \( -9k+2k-10=0 \).
3. So \( -7k=10 \) and \( k=-\frac{10}7 \).

Teacher's Note:
1. Perpendicular lines have a zero dot product of direction ratios.

Answer: (b) \( c:b \)

Explanation:
1. The XY-plane is \( z=0 \). Let it divide the segment in the ratio \( m:n \).
2. The z-coordinate of the dividing point is \( \frac{m(-b)+nc}{m+n}=0 \), so \( m:n=c:b \).
3. The ratio is \( c:b \).

Teacher's Note:
1. A point on the XY-plane has z = 0; apply the section formula to the z-coordinates.

Answer: (b) \( \frac{x-a}0=\frac{y-b}0=\frac{z-c}1 \)

Explanation:
1. A line parallel to the Z-axis has direction ratios \( (0,0,1) \).
2. Through \( (a,b,c) \), the symmetric form is \( \frac{x-a}0=\frac{y-b}0=\frac{z-c}1 \).
3. This is option d.

Teacher's Note:
1. The direction ratios of the Z-axis are 0, 0, 1.

Answer: (a) 0

Explanation:
1. Both lines pass through the point with position vector \( \vec a \) (put \( t=0 \) and \( s=0 \)).
2. Two lines with a common point intersect.
3. So the shortest distance is 0.

Teacher's Note:
1. Look for a common point before using the long formula.

Answer: (b) \(L\) lies in \( \pi \)

Explanation:
1. The direction of \(L\) is \( (3,2,1) \) and the normal of \( \pi \) is \( (1,-2,1) \). Their dot product is \( 3-4+1=0 \), so \(L\) is parallel to \( \pi \) or lies in it.
2. The point \( (1,-1,-3) \) of \(L\) satisfies \( 1+2-3=0 \), so it lies on \( \pi \).
3. Hence \(L\) lies in \( \pi \); this is stronger than being parallel, so b is the correct choice.

Teacher's Note:
1. Check both the direction and a point of the line.

Answer: (b) 2563

Explanation:
1. A general point of the line is \( (1+2\lambda,\ 2+\lambda,\ 3+2\lambda) \). Substituting in the first plane: \( 2(1+2\lambda)-6(2+\lambda)+3(3+2\lambda)+5=4+4\lambda=0 \), so \( \lambda=-1 \).
2. The point of intersection is \( (-1,1,1) \).
3. Then \( \alpha=-1+75+60=134 \) and \( 19\alpha+17=2546+17=2563 \).

Teacher's Note:
1. Find the parameter first, then the point, then alpha.

Answer: (d) \( 2x+3y-4z+7=0 \)

Explanation:
1. The normal is \( \vec{OP}=(2,3,-4) \) and the plane passes through \( \vec b=(2,-1,2) \).
2. The constant is \( 2(2)+3(-1)-4(2)=4-3-8=-7 \), so the plane is \( 2x+3y-4z=-7 \).
3. That is \( 2x+3y-4z+7=0 \).

Teacher's Note:
1. Substitute the given point to find the constant term.

Answer: (c) \( \frac{91}{12} \)

Explanation:
1. Writing the plane as \( 2x+3y-4z=-7 \), the intercepts are \( x=-\frac72,\ y=-\frac73,\ z=\frac74 \).
2. Their lengths are \( \frac72,\ \frac73,\ \frac74 \).
3. The sum is \( \frac{42+28+21}{12}=\frac{91}{12} \).

Teacher's Note:
1. Set two of the coordinates to zero to read off each intercept.

Answer: (a) 1624

Explanation:
1. \( \sin\theta=\frac{|3\cdot2+2\cdot1+4\cdot(-3)|}{\sqrt{9+4+16}\sqrt{4+1+9}}=\frac{4}{\sqrt{29\cdot14}} \).
2. \( \operatorname{cosec}^2\theta=\frac{406}{16} \).
3. \( 64\operatorname{cosec}^2\theta=406\times4=1624 \).

Teacher's Note:
1. The angle between a line and a plane uses sine, with the normal vector of the plane.

Answer: (c) meet in infinitely many points

Explanation:
1. The direction \( (1,2,3) \) and the normal \( (1,-2,1) \) have dot product \( 1-4+3=0 \), so the line is parallel to the plane.
2. The origin lies on both the line and the plane.
3. So the line lies in the plane and they meet in infinitely many points.

Teacher's Note:
1. A parallel line sharing one point with the plane lies entirely in it.

Answer: (d) \( (2,0,0) \)

Explanation:
1. The foot of the perpendicular on the X-axis keeps the x-coordinate and has y = z = 0.
2. The foot is \( (2,0,0) \).
3. Check: the vector \( (0,5,7) \) from the foot to the point is perpendicular to the X-axis.

Teacher's Note:
1. Projection on the X-axis only keeps the x-coordinate.

Answer: (c) \( n=0 \)

Explanation:
1. A line is parallel to the XY-plane when it is perpendicular to the Z-axis.
2. That requires its Z-direction ratio to vanish.
3. So \( n=0 \).

Teacher's Note:
1. Parallel to the XY-plane means the line has no z-component of direction.

Answer: (b) \( (-3,5,2) \)

Explanation:
1. For the plane \( 2x-y+z+3=0 \), \( 2(1)-3+4+3=6 \) and \( |\vec n|^2=6 \).
2. The image is \( P-\frac{2\cdot6}{6}(2,-1,1)=(1,3,4)-(4,-2,2) \).
3. So the image is \( (-3,5,2) \).

Teacher's Note:
1. Image = P - 2 (value of the plane at P) / |n|^2 times n.

Answer: (d) \( x+y+z-15=0 \)

Explanation:
1. The normal is along the joining vector \( (4,4,4)\parallel(1,1,1) \).
2. The plane passes through the midpoint \( (4,5,6) \): \( x+y+z=15 \).
3. So the plane is \( x+y+z-15=0 \).

Teacher's Note:
1. A perpendicular bisector plane has the segment as normal and passes through the midpoint.

Answer: (d) \( 3x^2-10x+7=0 \)

Explanation:
1. Distances are proportional to \( |3+4-12\lambda+13|=|20-12\lambda| \) and \( |-9+0-12+13|=8 \).
2. \( 20-12\lambda=\pm8 \) gives \( \lambda=1 \) or \( \lambda=\frac73 \).
3. The quadratic with these roots is \( 3\lambda^2-10\lambda+7=0 \).

Teacher's Note:
1. Equidistant means the absolute values of the plane expression are equal; the common denominator cancels.

Answer: (b) \( (0,7,10) \)

Explanation:
1. The normal is \( (1,5,4)\times(2,0,-3)=(-15,11,-10) \), where \( (2,0,-3) \) joins \( (1,2,3) \) to \( (3,2,0) \).
2. The plane is \( 15x-11y+10z=23 \), using the point \( (3,2,0) \).
3. Testing the options, only \( (0,7,10) \) gives \( -77+100=23 \).

Teacher's Note:
1. Use the point on the line and the given point to get a second direction in the plane.

Answer: (d) \( \left[\frac\pi4,\frac\pi2\right] \)

Explanation:
1. With direction cosines \( \cos\theta,\cos\theta,\cos\gamma \): \( 2\cos^2\theta+\cos^2\gamma=1 \).
2. So \( 2\cos^2\theta\le1 \), i.e. \( \cos\theta\le\frac1{\sqrt2} \), giving \( \theta\ge\frac\pi4 \).
3. Together with \( \theta\le\frac\pi2 \), \( \theta\in\left[\frac\pi4,\frac\pi2\right] \).

Teacher's Note:
1. The squared direction cosine of the third axis must be non-negative.

Answer: (d) 4

Explanation:
1. The normal of the plane is \( \vec n\propto(7,-11,1) \). The plane passes through \( (2,-1,3) \), a point of the first line.
2. \( 7(2)-11(-1)+3=14+11+3=28 \).
3. So \( 7k=28 \) and \( k=4 \).

Teacher's Note:
1. The constant term follows from one known point on the plane.

Answer: (b) 3

Explanation:
1. The YZ-plane is \( x=0 \). If it divides the segment in the ratio \( m:n \), then \( \frac{2m-3n}{m+n}=0 \).
2. So \( m:n=3:2 \).
3. Hence \( k=3 \).

Teacher's Note:
1. A point on the YZ-plane has x = 0.

Answer: (a) 1

Explanation:
1. The angle \( \theta \) between the line and the plane has \( \cos\theta=\sqrt{\frac5{14}} \), so \( \sin\theta=\sqrt{\frac9{14}} \).
2. \( \sin\theta=\frac{|1+4+3\lambda|}{\sqrt{5+\lambda^2}\sqrt{14}} \). Squaring: \( (5+3\lambda)^2=9\left(5+\lambda^2\right) \), so \( 25+30\lambda=45 \).
3. \( \lambda=\frac23 \), so \( l=2,\ m=3 \) and \( m-l=1 \).

Teacher's Note:
1. For a line and a plane the sine of the angle uses the dot product with the normal.

Answer: (b) 5

Explanation:
1. The direction \( (3,1,-5) \) is perpendicular to the normal \( (2,-1,1) \) (\( 6-1-5=0 \)), so the line is parallel to the plane and its image has the same direction.
2. The image of the point \( (1,3,4) \) is \( (1,3,4)-\frac{2\cdot6}6(2,-1,1)=(-3,5,2) \).
3. So the image line passes through \( (-3,5,2) \), giving \( k=5 \).

Teacher's Note:
1. A line parallel to the mirror plane has an image with the same direction through the image of any of its points.