JEE Mathematics Three Dimensional Geometry MCQs Set 08

Answer: (b) x + 2y + 3z = 14

Explanation:
1. The normal is (1, 2, 3).
2. The plane is \( x+2y+3z=k \) through (1, 2, 3).
3. So k = 1 + 4 + 9 = 14.

Teacher's Note:
1. The foot gives both a point of the plane and its normal direction.
2. Substitute the foot into x + 2y + 3z = k.

Answer: (c) (3, 2, 1)

Explanation:
1. Substitute: \( 2(3)-2+3(1)=7 \).
2. The point already lies on the plane.
3. The image of a point on the mirror is the point itself.

Teacher's Note:
1. Always test whether the point lies on the mirror plane before using the image formula.
2. If it does, the image coincides with the point.

Answer: (d) 2x - y + 2z + 1 = 0

Explanation:
1. The normal is the vector joining the points: (4, 2, 1) - (2, 3, -1) = (2, -1, 2).
2. Plane: \( 2x-y+2z=k \) through (2, 3, -1).
3. \( k=4-3-2=-1 \), so \( 2x-y+2z+1=0 \).

Teacher's Note:
1. The line from the external point to the foot is the normal.
2. Use the foot as the point on the plane.

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

Explanation:
1. Equal distances: \( |3+4+12p+13|=|-9+0+12+13| \).
2. So \( |12p+20|=16 \), giving \( 12p+20=\pm16 \).
3. Hence \( p=-\frac13 \) or p = -3.

Teacher's Note:
1. Equidistance means the absolute values of the plane expression are equal.
2. Solve both signs and match with the options.

Answer: (c) -1 : 4

Explanation:
1. At the first point: \( 6-8+3-4=-3 \); at the second: \( -12+10-6-4=-12 \).
2. Ratio \( =-\frac{-3}{-12}=-\frac14 \).
3. The plane divides the line externally in the ratio -1 : 4.

Teacher's Note:
1. Ratio = -(value at first) / (value at second).
2. A negative ratio means external division.

Answer: (b) lie on opposite sides of the plane

Explanation:
1. At A: \( -8-3+4-3=-10 \).
2. At B: \( 12-3-4-3=2 \).
3. The signs differ, so the points are on opposite sides.

Teacher's Note:
1. Compare the signs of the plane expression at the two points.
2. Different signs mean opposite sides.

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

Explanation:
1. The d.r.'s are (2, -1, 2).
2. Its length is \( \sqrt{4+1+4}=3 \).
3. The d.c.'s are \( \left(\frac23,-\frac13,\frac23\right) \) (or the negatives).

Teacher's Note:
1. Direction cosines are the d.r.'s divided by the length of the vector.
2. Check that the squares add up to 1.

Answer: (a) 3

Explanation:
1. \( \frac{|-9|}{\sqrt{1+4+4}}=\frac93=3 \).

Teacher's Note:
1. The distance from the origin is |d| / sqrt(a^2 + b^2 + c^2).
2. Here sqrt(1 + 4 + 4) = 3.

Answer: (b) \( 3\sqrt{21} \)

Explanation:
1. Numerator: \( |14+56-5-2|=63 \).
2. Denominator: \( \sqrt{4+16+1}=\sqrt{21} \).
3. Length \( =\frac{63}{\sqrt{21}}=3\sqrt{21} \).

Teacher's Note:
1. Use |ax1 + by1 + cz1 + d| / sqrt(a^2 + b^2 + c^2).
2. Rationalise: 63/sqrt(21) = 3 sqrt(21).

Answer: (d) 7y + 5z = 35

Explanation:
1. A plane parallel to the x-axis has the form \( \frac y5+\frac z7=1 \).
2. Multiplying by 35: \( 7y+5z=35 \).

Teacher's Note:
1. Parallel to the x-axis means x does not appear.
2. Use intercept form for the other two variables and clear fractions.

Answer: (a) 15, -10, 6

Explanation:
1. Write it as \( 2x-3y+5z=30 \).
2. Divide by 30: \( \frac x{15}+\frac y{-10}+\frac z6=1 \).
3. The intercepts are 15, -10, 6.

Teacher's Note:
1. Put the equation in intercept form by dividing by the constant.
2. Signs of the intercepts follow the signs of the coefficients.

Answer: (c) 6

Explanation:
1. The plane cuts the X-axis at (4, 0, 0) and the Y-axis at (0, 3, 0).
2. The triangle with the origin has area \( \frac12\cdot4\cdot3=6 \).

Teacher's Note:
1. Only the x- and y-intercepts matter.
2. The z-intercept is not needed.

Answer: (d) 6

Explanation:
1. Parallel planes have proportional coefficients: \( \frac24=\frac3k=\frac48 \).
2. So k = 6.

Teacher's Note:
1. Equate the ratios of the coefficients of x, y, z.
2. The constant terms are ignored.

Answer: (b) parallel

Explanation:
1. Ratios of coefficients: \( \frac25=\frac{-1}{-2.5}=\frac4{10}=0.4 \).
2. The coefficients are proportional, but the constants are not.
3. So the planes are parallel (and distinct).

Teacher's Note:
1. Compare the coefficient ratios first.
2. Equal ratios with a different constant ratio mean distinct parallel planes.

Answer: (b) 2

Explanation:
1. Rewrite the second as \( 2x+2\lambda y+5z=7 \).
2. Matching with the first: \( \lambda=2 \) and \( 2\lambda=4 \).
3. Both give lambda = 2.

Teacher's Note:
1. Make the right-hand sides and the z-coefficients equal first.
2. Then compare the x and y coefficients.

Answer: (d) a + c + d - 4 = 0

Explanation:
1. Perpendicular to the ZX-plane means the plane contains the y-direction, so b = 0.
2. Substituting the points: a + 3c + d = 0 and -a + d = 0.
3. Taking a = 3: d = 3, c = -2, so the plane is 3x - 2z + 3 = 0, and a + c + d - 4 = 3 - 2 + 3 - 4 = 0.

Teacher's Note:
1. A plane perpendicular to a coordinate plane has no coefficient for the third variable.
2. Use the two points to find the remaining ratios.

Answer: (b) \( (ab'-a'b)y+(ac'-a'c)z+(ad'-a'd)=0 \)

Explanation:
1. The line y = 0 = z is the x-axis, with d.r.'s (1, 0, 0).
2. Use \( P+\lambda P'=0 \) and require the coefficient of x to vanish: \( a+\lambda a'=0 \), so \( \lambda=-a/a' \).
3. Then \( a'P-aP'=0 \) gives \( (a'b-ab')y+(a'c-ac')z+(a'd-ad')=0 \), which is option 2 multiplied by -1.

Teacher's Note:
1. A plane parallel to a line has a normal perpendicular to that line's direction.
2. Here that kills the x-coefficient.

Answer: (b) \( 1,1,\sqrt2 \)

Explanation:
1. Let the normal be (l, m, n). Using the points: l(x - 1) + my + nz = 0 passes through (0, 1, 0), so m = l.
2. Angle with x + y = 3: \( \cos\frac\pi4=\frac{|l+m|}{\sqrt2\sqrt{l^2+m^2+n^2}} \).
3. With m = l this gives \( 4l^2=2l^2+n^2 \), so \( n=\sqrt2\,l \): (1, 1, sqrt 2).

Teacher's Note:
1. Use both points to relate l and m.
2. Then use the angle formula for a plane with another plane.

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

Explanation:
1. AB = (1, 2, 4), AC = (-2, -1, 1); the cross product is (6, -9, 3), i.e. (2, -3, 1).
2. Plane ABC: 2x - 3y + z = 0.
3. A parallel plane at distance 2: \( \frac{|k|}{\sqrt{14}}=2 \), so \( k=\pm2\sqrt{14} \).

Teacher's Note:
1. Find the normal by the cross product of two edge vectors.
2. Then fix the constant by the distance condition.

Answer: (c) \( \frac1{x^2}+\frac1{y^2}+\frac1{z^2}=\frac{16}{p^2} \)

Explanation:
1. Plane \( \frac xa+\frac yb+\frac zc=1 \) at distance p: \( \frac1{a^2}+\frac1{b^2}+\frac1{c^2}=\frac1{p^2} \).
2. Centroid of OABC is (a/4, b/4, c/4), so a = 4x, b = 4y, c = 4z.
3. Substituting: \( \frac1{16}\left(\frac1{x^2}+\frac1{y^2}+\frac1{z^2}\right)=\frac1{p^2} \).

Teacher's Note:
1. The centroid of a tetrahedron with one vertex at the origin has one fourth of the intercepts.
2. Substitute into the distance condition.

Answer: (c) x + 2y + 2z - 6 = 0

Explanation:
1. Take \( (2+\lambda)x+(1-\lambda)y+(3+\lambda)z+(4\lambda-2)=0 \).
2. Distance 2: \( (4\lambda-2)^2=4\left[(2+\lambda)^2+(1-\lambda)^2+(3+\lambda)^2\right] \), i.e. \( \lambda^2-12\lambda-13=0 \).
3. So lambda = -1 or 13. lambda = -1 gives x + 2y + 2z - 6 = 0 (lambda = 13 gives 15x - 12y + 16z + 50 = 0).

Teacher's Note:
1. Write the family of planes through the line and impose the distance condition.
2. The quadratic in lambda has two roots, giving two planes.

Answer: (d) 8y + 3z = 24

Explanation:
1. Parallel to the x-axis means \( \frac y3+\frac z8=1 \).
2. Multiplying by 24: \( 8y+3z=24 \).

Teacher's Note:
1. Use intercept form for y and z only.
2. Multiply by the product of the intercepts to clear fractions.

Answer: (a) 2

Explanation:
1. Mid-point (1, 2, 1); normal AB = (2, 2, -2), i.e. (1, 1, -1).
2. Plane: \( x+y-z=1+2-1=2 \).
3. Intercepts: 2, 2, -2, whose sum is 2.

Teacher's Note:
1. The plane through the mid-point with the segment as normal is the perpendicular bisector plane.
2. Then read the intercepts.