JEE Mathematics Introduction to Three Dimensional Geometry MCQs Set 03

Answer: (a) x-axis

Explanation:
1. The \(yz\)-plane is \(x=0\), so a parallel plane has the form \(x=k\).
2. Its normal is along the x-axis, so it is perpendicular to the x-axis.

Teacher's Note:
1. A plane parallel to a coordinate plane is perpendicular to the third axis.
2. The yz-plane has equation x = 0.

Answer: (a) \((0,-6,0)\text{ or }(0,2,0)\)

Explanation:
1. Let the point be \((0,y,0)\). Then \(9+(y+2)^2+25=50\).
2. So \(y^2+4y-12=0\), giving \(y=-6\) or \(y=2\).

Teacher's Note:
1. Points on the y-axis have x = z = 0.
2. Square both sides to remove the root.

Answer: (b) \((-3,7,2)\text{ and }(2,4,-1)\)

Explanation:
1. Option (b): \(\sqrt{25+9+9}=\sqrt{43}\).
2. The other pairs give \(\sqrt{20},\ \sqrt{104},\ \sqrt{20}\).

Teacher's Note:
1. Compute each squared distance and compare with 43.
2. Use the distance and section formulas directly and check the arithmetic with a quick substitution.

Answer: (b) \(x-2z=0\)

Explanation:
1. Equate squared distances: \((x-1)^2+(z-3)^2=(x-3)^2+(z+1)^2\) (the y terms cancel).
2. Expanding gives \(-2x-6z+10=-6x+2z+10\), so \(4x-8z=0\), i.e. \(x-2z=0\).

Teacher's Note:
1. The locus of points equidistant from two points is their perpendicular bisector plane.
2. Cancel common squared terms before expanding.

Answer: (a) collinear

Explanation:
1. The vectors are \(\overrightarrow{AB}=(-3,-1,-7)\) and \(\overrightarrow{AC}=(-6,-2,-14)=2\overrightarrow{AB}\).
2. They are parallel and share the point A, so the points are collinear (the triangle area is zero).

Teacher's Note:
1. Three points are collinear when two joining vectors are parallel.
2. Equivalent test: the area of the triangle is zero.

Answer: (b) \((-1,6,6)\)

Explanation:
1. Compute \(AB^2=18,\ BC^2=18,\ AC^2=36\).
2. So \(AB^2+BC^2=AC^2\) and the triangle is right angled at B.
3. The orthocentre of a right triangle is the vertex at the right angle, \(B(-1,6,6)\).

Teacher's Note:
1. In a right triangle the orthocentre is the right-angle vertex.
2. Check the Pythagoras relation on the squared lengths.

Answer: (b) \(\left(\frac{13}3,\frac13,-\frac53\right)\)

Explanation:
1. The sides are \(AB=BC=CA=\sqrt{72}\), so the triangle is equilateral.
2. In an equilateral triangle the circumcentre equals the centroid \(\left(\frac{1+3+9}3,\frac{-1+1+1}3,\frac{-5+3-3}3\right)=\left(\frac{13}3,\frac13,-\frac53\right)\).

Teacher's Note:
1. For an equilateral triangle all four centres coincide.
2. Average the coordinates to get the centroid.

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

Explanation:
1. The diagonals bisect each other, so \(A+C=B+D\).
2. Hence \(D=A+C-B=(3-1-1,\,-1+1-2,\,2+2+4)=(1,-2,8)\).

Teacher's Note:
1. In a parallelogram ABCD, the sum of opposite vertices is equal.
2. D = A + C - B.

Answer: (c) \(\frac{\sqrt{155}}2\)

Explanation:
1. The vectors are \((1,3,-4)\) and \((-2,-1,-1)\).
2. Cross product: \((-7,\,9,\,5)\), with magnitude \(\sqrt{155}\).
3. Area \(=\frac12\sqrt{155}\).

Teacher's Note:
1. Area of a triangle in space = half the magnitude of the cross product.
2. Compute two edge vectors from one vertex.

Answer: (d) eighth octant

Explanation:
1. All three coordinates are negative.
2. The octant with signs \((-,-,-)\) is the eighth octant.

Teacher's Note:
1. The eighth octant has all coordinates negative.
2. The first octant has all positive.

Answer: (d) None of these

Explanation:
1. The foot on the \(xy\)-plane keeps x and y and has z = 0.
2. So \(L=(3,4,0)\), which is not listed.

Teacher's Note:
1. The foot of the perpendicular on the xy-plane has z = 0.
2. Keep the other two coordinates.

Answer: (a) 3 units

Explanation:
1. The foot on the \(yz\)-plane is \((0,4,5)\).
2. The distance is the absolute value of the x-coordinate, 3.

Teacher's Note:
1. The distance from the yz-plane is |x|.
2. Similarly |y| for the zx-plane and |z| for the xy-plane.

Answer: (b) \(\sqrt{34}\)

Explanation:
1. The foot on the y-axis is \((0,4,0)\).
2. Distance \(=\sqrt{3^2+0+5^2}=\sqrt{34}\).

Teacher's Note:
1. The foot of a perpendicular on the y-axis keeps only the y coordinate.
2. Then use the distance formula.

Answer: (b) \(-1\)

Explanation:
1. Let \(P\) divide \(QR\) in the ratio \(\lambda:1\). Then \(\frac{5\lambda+2}{\lambda+1}=4\) gives \(\lambda=2\).
2. So \(z=\frac{2(-2)+1}{3}=-1\).

Teacher's Note:
1. Find the ratio from the given x-coordinate first.
2. Apply the same ratio to z.

Answer: (b) \(2(x^2+y^2+z^2)-4x-14y+4z+109-K^2=0\)

Explanation:
1. Write \((x-3)^2+(y-4)^2+(z-5)^2+(x+1)^2+(y-3)^2+(z+7)^2=K^2\).
2. Expanding: \(2x^2+2y^2+2z^2-4x-14y+4z+109=K^2\) since \(9+16+25+1+9+49=109\).

Teacher's Note:
1. Expand each squared distance and add the constants.
2. Check the constant: 9+16+25+1+9+49 = 109.

Answer: (b) \(\sqrt{13},\sqrt{10},\sqrt5\)

Explanation:
1. Distance from the x-axis is \(\sqrt{y^2+z^2}=\sqrt{13}\), from the y-axis \(\sqrt{x^2+z^2}=\sqrt{10}\), from the z-axis \(\sqrt{x^2+y^2}=\sqrt5\).

Teacher's Note:
1. The distance from an axis ignores the coordinate along that axis.
2. Keep the order x-axis, y-axis, z-axis.

Answer: (a) \(\left(\frac a2,\frac b2,\frac c2\right)\)

Explanation:
1. Equal distance from the origin and \((a,0,0)\) gives \(x=\frac a2\); similarly \(y=\frac b2\) and \(z=\frac c2\).

Teacher's Note:
1. Equidistance from the origin and a point on an axis fixes that coordinate at half.
2. Do it axis by axis.

Answer: (a) 1

Explanation:
1. The distance is \(\sqrt{x^2+y^2+1-x^2-y^2}=1\).

Teacher's Note:
1. Substitute the third coordinate and simplify.
2. The point lies on the unit sphere.

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

Explanation:
1. The edges are \(|5-3|=2,\ |8-6|=2,\ |10-8|=2\).
2. The diagonal is \(\sqrt{4+4+4}=2\sqrt3\).

Teacher's Note:
1. The diagonal of a box is the root of the sum of squares of edges.
2. The two points are opposite corners.

Answer: (b) internally 2 : 3

Explanation:
1. On the \(YZ\)-plane \(x=0\): \(\frac{3k-2}{k+1}=0\) gives \(k=\frac23\).

Teacher's Note:
1. Set the x-coordinate of the dividing point to zero.
2. A positive ratio means internal division.

Answer: (a) \(7,7,\sqrt{34}\)

Explanation:
1. Midpoints: \(BC:(3,2,0),\ CA:(3,0,3),\ AB:(0,2,3)\).
2. Medians: \(\sqrt{9+4+36}=7\), \(\sqrt{9+16+9}=\sqrt{34}\), \(\sqrt{36+4+9}=7\).

Teacher's Note:
1. A median joins a vertex to the midpoint of the opposite side.
2. Compute three distances.

Answer: (b) \((6,-4,-2),(8,-10,2)\)

Explanation:
1. Use the ratios 1 : 2 and 2 : 1: \(\frac{2(4,2,-6)+(10,-16,6)}{3}=(6,-4,-2)\) and \(\frac{(4,2,-6)+2(10,-16,6)}{3}=(8,-10,2)\).

Teacher's Note:
1. Trisection points divide in the ratios 1 : 2 and 2 : 1.
2. Keep track of which end gets weight 2.

Answer: (b) rhombus

Explanation:
1. All four sides equal 7: \(AB=BC=CD=DA=7\).
2. The diagonals are \(AC=\sqrt{122}\) and \(BD=\sqrt{74}\), which are unequal, so it is not a square.

Teacher's Note:
1. Equal sides with unequal diagonals mean a rhombus.
2. Check the diagonals to rule out a square.

Answer: (c) a parallelogram

Explanation:
1. The sums \((5,-4,2)+(7,-6,4)\) and \((4,-3,1)+(8,-7,5)\) are both \((12,-10,6)\), so the diagonals bisect each other.
2. Sides are \(\sqrt3\) and \(3\sqrt3\), diagonals \(2\sqrt3\) and \(4\sqrt3\), which are unequal, so it is a parallelogram and not a rectangle.

Teacher's Note:
1. Equal midpoints of diagonals give a parallelogram.
2. Unequal diagonals rule out a rectangle.

Answer: (c) 90°

Explanation:
1. \(\overrightarrow{AB}=(3,0,3)\) and \(\overrightarrow{AC}=(4,2,-4)\).
2. Dot product \(12+0-12=0\), so the angle is 90°.

Teacher's Note:
1. Zero dot product means perpendicular vectors.
2. Use vectors from A.

Answer: (b) \(2\sqrt6\)

Explanation:
1. \(\overrightarrow{AB}=(1,1,-3)\), \(\overrightarrow{AC}=(-1,3,-1)\).
2. Cross product \((8,4,4)\); magnitude \(\sqrt{96}=4\sqrt6\).
3. Area \(=\frac12\cdot4\sqrt6=2\sqrt6\).

Teacher's Note:
1. Area = half the magnitude of AB x AC.
2. Take the cross product component by component.

Answer: (d) right angled isosceles

Explanation:
1. Squared sides are 18, 18 and 36.
2. Two equal sides and \(18+18=36\) give a right-angled isosceles triangle.

Teacher's Note:
1. Check equal sides, then the Pythagoras relation.
2. Both properties hold, so choose the most specific.

Answer: (a) collinear

Explanation:
1. \(\overrightarrow{AB}=(1,-3,2)\) and \(\overrightarrow{AC}=(4,-12,8)=4\,\overrightarrow{AB}\).
2. So they are collinear.

Teacher's Note:
1. Parallel joining vectors from the same point show collinearity.
2. Check that all three ratios are equal.

Answer: (b) \(\left(\frac73,\frac53,\frac53\right)\)

Explanation:
1. The centroid divides the join of orthocentre \(H\) and circumcentre \(S\) in the ratio \(2:1\) from \(H\).
2. So \(G=\frac{H+2S}{3}=\left(\frac{1+6}3,\frac{1+4}3,\frac{1+4}3\right)=\left(\frac73,\frac53,\frac53\right)\).

Teacher's Note:
1. Euler line: HG : GS = 2 : 1.
2. G = (H + 2S) / 3.

Answer: (b) \((2,3,-1)\)

Explanation:
1. \(AB=3,\ BC=3,\ CA=\sqrt{18}\) and \(AB^2+BC^2=CA^2\).
2. The triangle is right angled at B, so the orthocentre is \(B(2,3,-1)\).

Teacher's Note:
1. The orthocentre of a right triangle is the right-angle vertex.
2. Verify with squared side lengths.