JEE Mathematics Three Dimensional Geometry MCQs Set 07

Answer: (b) (0, 0, 0)

Explanation:
1. The normal is (6, 20, -1), so the plane is \( 6(x+1)+20(y-0)-(z+6)=0 \).
2. This simplifies to \( 6x+20y-z=0 \).
3. Only (0, 0, 0) satisfies it; for (1, 1, -26) the value is 52, not 0.

Teacher's Note:
1. A plane through a point with a known normal is a(x - x1) + b(y - y1) + c(z - z1) = 0.
2. Substitute each option to see which one satisfies the equation.

Answer: (b) \( 5x+3y-11z=\frac{135}{2} \)

Explanation:
1. Mid-point of AB is \( \left(\frac32,\frac72,-\frac92\right) \) and AB has d.r.'s (5, 3, -11).
2. The plane is \( 5x+3y-11z=5\cdot\frac32+3\cdot\frac72-11\cdot\left(-\frac92\right) \).
3. The right side is \( \frac{15+21+99}{2}=\frac{135}{2} \).

Teacher's Note:
1. The segment AB itself serves as the normal of the plane.
2. The plane must pass through the mid-point, so substitute it to find the constant.

Answer: (a) YOZ

Explanation:
1. The normal of the plane is (0, b, c), which has no x-component.
2. So the plane is parallel to the x-axis.
3. A plane parallel to the x-axis is perpendicular to the YOZ plane.

Teacher's Note:
1. A missing variable means the plane is parallel to that axis.
2. Its normal then lies in the coordinate plane of the other two variables.

Answer: (b) 3x+4y+5z-26=0

Explanation:
1. A parallel plane has the same normal (3, 4, 5): \( 3x+4y+5z=k \).
2. Through (1, 2, 3): \( k=3+8+15=26 \).
3. Hence \( 3x+4y+5z-26=0 \).

Teacher's Note:
1. Parallel planes share the same coefficients of x, y, z.
2. Only the constant term changes.

Answer: (c) \( \frac72 \)

Explanation:
1. Rewrite the second plane as \( 2x+y+2z+\frac52=0 \).
2. The first is \( 2x+y+2z-8=0 \).
3. Distance \( =\frac{|\frac52-(-8)|}{\sqrt{4+1+4}}=\frac{21/2}{3}=\frac72 \).

Teacher's Note:
1. Make the coefficients of x, y, z identical before using the distance formula.
2. Then the distance is |d1 - d2| divided by the length of the normal.

Answer: (b) x+2y+2z-9=0

Explanation:
1. The normal is the vector from the origin to the foot, (1, 2, 2).
2. The plane is \( x+2y+2z=k \) and passes through (1, 2, 2).
3. So \( k=1+4+4=9 \).

Teacher's Note:
1. The line from the origin to the foot is perpendicular to the plane.
2. So the foot's coordinates give the normal directly.

Answer: (b) (1, 2, 8)

Explanation:
1. Value of the plane expression at P: \( 14+56-5-2=63 \) and \( a^2+b^2+c^2=21 \).
2. Foot \( =P-\frac{63}{21}(2,4,-1)=(7-6,14-12,5+3) \).
3. The foot is (1, 2, 8).

Teacher's Note:
1. Foot = P - [(ax1 + by1 + cz1 + d)/(a^2 + b^2 + c^2)] times (a, b, c).
2. Verify that the result satisfies the plane equation.

Answer: (b) \( \left(\frac95,-\frac{13}5,4\right) \)

Explanation:
1. Value at P: \( -1-6=-7 \), and \( a^2+b^2+c^2=5 \).
2. Image \( =P-2\cdot\frac{-7}{5}(1,-2,0)=\left(-1+\frac{14}5,\,3-\frac{28}5,\,4\right) \).
3. The image is \( \left(\frac95,-\frac{13}5,4\right) \).

Teacher's Note:
1. Image = P - 2[(ax1 + by1 + cz1 + d)/(a^2 + b^2 + c^2)] times (a, b, c).
2. The z-coordinate does not change because c = 0.

Answer: (a) opposite sides

Explanation:
1. At (1, -1, 1): \( 2-3-1+7=5>0 \).
2. At (-2, 0, 5): \( -4+0-5+7=-2<0 \).
3. The values have opposite signs, so the points lie on opposite sides.

Teacher's Note:
1. Substitute each point in ax + by + cz + d.
2. Different signs mean opposite sides; equal signs mean the same side.

Answer: (d) 3 : 10

Explanation:
1. The plane is \( x-2y+3z-17=0 \).
2. At the first point: \( -2-8+21-17=-6 \); at the second: \( 3+10+24-17=20 \).
3. The ratio is \( -(-6):20=6:20=3:10 \).

Teacher's Note:
1. The ratio is -(value at first point) : (value at second point).
2. A positive ratio means the plane cuts the segment internally.

Answer: (a) on the side opposite to the origin

Explanation:
1. At the point: \( 4+9-25+10=-2<0 \).
2. At the origin the value is 10 > 0.
3. The signs differ, so the point lies on the side opposite to the origin.

Teacher's Note:
1. Compare the sign of the plane expression at the point with its sign at the origin.
2. Opposite signs mean opposite sides.

Answer: (d) \( \frac23x-\frac23y+\frac13z=\frac53 \)

Explanation:
1. Here \( \sqrt{a^2+b^2+c^2}=\sqrt{4+4+1}=3 \).
2. Divide the equation by 3.
3. This gives \( \frac23x-\frac23y+\frac13z=\frac53 \), with positive right side.

Teacher's Note:
1. Divide by the length of the normal so that the coefficients become direction cosines.
2. The right side then equals the distance from the origin, which must be positive.

Answer: (d) \( \frac{13}{\sqrt{74}} \)

Explanation:
1. The plane is \( 3(x-2)-4(y-3)+7(z+1)=0 \), i.e. \( 3x-4y+7z+13=0 \).
2. Distance from the origin \( =\frac{13}{\sqrt{9+16+49}} \).
3. This equals \( \frac{13}{\sqrt{74}} \).

Teacher's Note:
1. First find the plane from the point and the normal.
2. Then the distance from the origin is |d| divided by the length of the normal.

Answer: (c) x-2y+2z-3=0

Explanation:
1. A parallel plane is \( x-2y+2z+k=0 \).
2. Its distance from the origin is \( \frac{|k|}{3}=1 \), so \( k=\pm3 \).
3. Among the options, \( k=-3 \) appears.

Teacher's Note:
1. Keep the coefficients and vary the constant.
2. Equate |k| / sqrt(a^2 + b^2 + c^2) to the required distance.

Answer: (b) x + y + z = 1

Explanation:
1. Intercept form: \( \frac x a+\frac y b+\frac z c=1 \) with a = b = c = 1.
2. So the plane is x + y + z = 1.

Teacher's Note:
1. Use the intercept form directly.
2. Unit intercepts mean a = b = c = 1.

Answer: (a) (143, 91, 77)

Explanation:
1. Intercepts: \( A=\left(\frac{3003}7,0,0\right)=(429,0,0) \), B = (0, 273, 0), C = (0, 0, 231).
2. Centroid \( =\left(\frac{429}3,\frac{273}3,\frac{231}3\right) \).

Teacher's Note:
1. Find the three intercepts by setting two variables to zero.
2. The centroid is the average of the three vertices.

Answer: (a) \( \frac x4+\frac y6+\frac z3=1 \)

Explanation:
1. \( \tfrac12ab=12,\ \tfrac12bc=9,\ \tfrac12ca=6 \), so ab = 24, bc = 18, ca = 12.
2. Multiplying, \( (abc)^2=5184 \), so abc = 72.
3. Hence a = 72/18 = 4, b = 72/12 = 6, c = 72/24 = 3.

Teacher's Note:
1. Each triangle area is half the product of two intercepts.
2. Multiply the three products to get abc, then divide.

Answer: (d) 2

Explanation:
1. Perpendicular planes have perpendicular normals: \( a_1a_2+b_1b_2+c_1c_2=0 \).
2. \( 1(2)+2(1)+k(-2)=0 \), so \( 4-2k=0 \).
3. k = 2.

Teacher's Note:
1. Planes are perpendicular when the dot product of their normals is zero.
2. Substitute and solve the linear equation.

Answer: (d) \( \frac\pi3 \)

Explanation:
1. \( \cos\theta=\frac{|2-1+2|}{\sqrt6\,\sqrt6}=\frac36=\frac12 \).
2. So \( \theta=\frac\pi3 \).

Teacher's Note:
1. The angle between planes equals the angle between their normals.
2. Use the absolute value to get the acute angle.

Answer: (d) -8

Explanation:
1. Condition: \( 2(1)+3(2)+(-1)(-k)=0 \).
2. This gives \( 2+6+k=0 \).
3. So k = -8.

Teacher's Note:
1. Perpendicular planes have perpendicular normals.
2. Be careful with the double negative in the z-term.

Answer: (a) 6

Explanation:
1. Vectors in the plane: (-4, 2, 0) and (-3, 0, 1).
2. Their cross product is (2, 4, 6), i.e. d.r.'s (1, 2, 3).
3. The product is 1 x 2 x 3 = 6.

Teacher's Note:
1. A line perpendicular to the plane is parallel to its normal.
2. Obtain the normal as the cross product of two vectors lying in the plane.

Answer: (a) x - y + z = 1

Explanation:
1. The point (3, 2, 0) lies on the line (t = -1 gives (3, 2, 0)), so the plane only needs to contain the line.
2. The plane must contain (4, 7, 4) and be parallel to (1, 5, 4).
3. For x - y + z = 1: 4 - 7 + 4 = 1 and 1 - 5 + 4 = 0, and (3, 2, 0) also satisfies it.

Teacher's Note:
1. Check each option for the line's point and direction.
2. The normal must be perpendicular to the direction (1, 5, 4).

Answer: (a) 2x+3y+4z-9=0

Explanation:
1. A parallel plane is \( 2x+3y+4z+k=0 \).
2. Through (1, 1, 1): \( 2+3+4+k=0 \), so k = -9.

Teacher's Note:
1. Parallel planes keep the same coefficients.
2. Substitute the point to find the new constant.

Answer: (d) x - 3 = 0

Explanation:
1. A plane perpendicular to the x-axis has normal (1, 0, 0).
2. So it is x = constant.
3. Through (3, -6, 9) the constant is 3, giving x - 3 = 0.

Teacher's Note:
1. The normal of the plane is along the axis it is perpendicular to.
2. Then only that coordinate appears in the equation.

Answer: (b) 1

Explanation:
1. Rewrite the second as \( 7x+4y-4z-6=0 \).
2. Distance \( =\frac{|3-(-6)|}{\sqrt{49+16+16}}=\frac99 \).
3. The distance is 1.

Teacher's Note:
1. Divide the second equation by 2 so that the coefficients match.
2. Then apply |d1 - d2| / sqrt(a^2 + b^2 + c^2).