OP Malhotra Class 11 Maths Solutions Chapter 26 Points and their Coordinates in 3 Dimensions Exercise 26 (A)

Question 1. Find the distance from the origin to each of the points :
(i) (2, 2, 3)
(ii) (4, -1, 2)
(iii) (0, 4, -4)
(iv) (-4, -3, -2)
Answer:
(i) To find the distance from the origin O(0, 0, 0) to point P(2, 2, 3), we use the distance formula. We subtract the coordinates of the origin from the point, square them, add them up, and then take the square root.
Required distance \( = \sqrt{(2-0)^2+(2-0)^2+(3-0)^2} \)
\( = \sqrt{2^2+2^2+3^2} \)
\( = \sqrt{4+4+9} \)
\( = \sqrt{17} \)
(ii) For point (4, -1, 2), we do the same calculation.
Required distance \( = \sqrt{(4-0)^2+(-1-0)^2+(2-0)^2} \)
\( = \sqrt{4^2+(-1)^2+2^2} \)
\( = \sqrt{16+1+4} \)
\( = \sqrt{21} \)
(iii) For point (0, 4, -4), we apply the distance formula again.
Required distance \( = \sqrt{(0-0)^2+(4-0)^2+(-4-0)^2} \)
\( = \sqrt{0^2+4^2+(-4)^2} \)
\( = \sqrt{0+16+16} \)
\( = \sqrt{32} \)
\( = 4\sqrt{2} \)
(iv) For point (-4, -3, -2), we calculate the distance from the origin.
Required distance \( = \sqrt{(-4-0)^2+(-3-0)^2+(-2-0)^2} \)
\( = \sqrt{(-4)^2+(-3)^2+(-2)^2} \)
\( = \sqrt{16+9+4} \)
\( = \sqrt{29} \)
In simple words: To find how far a point is from the starting point (origin), we square each of its coordinates, add those squares together, and then find the square root of the total.

🎯 Exam Tip: Remember the 3D distance formula from the origin is \( \sqrt{x^2+y^2+z^2} \). Always show each step of squaring and adding to avoid calculation errors.

Question 2. Find the distance between each of the following pairs of points :
(i) (2, 5, 3) and (-3, 2, 1);
(ii) (0, 3, 0) and (6, 0, 2);
(iii) (-4, -2, 0) and (3, 3, 5).
Answer:
(i) To find the distance between two points \( (x_1, y_1, z_1) \) and \( (x_2, y_2, z_2) \), we use the 3D distance formula. For (2, 5, 3) and (-3, 2, 1):
Required distance \( = \sqrt{(-3-2)^2+(2-5)^2+(1-3)^2} \)
\( = \sqrt{(-5)^2+(-3)^2+(-2)^2} \)
\( = \sqrt{25+9+4} \)
\( = \sqrt{38} \)
(ii) For (0, 3, 0) and (6, 0, 2):
Required distance \( = \sqrt{(6-0)^2+(0-3)^2+(2-0)^2} \)
\( = \sqrt{6^2+(-3)^2+2^2} \)
\( = \sqrt{36+9+4} \)
\( = \sqrt{49} \)
\( = 7 \)
(iii) For (-4, -2, 0) and (3, 3, 5):
Required distance \( = \sqrt{(3-(-4))^2+(3-(-2))^2+(5-0)^2} \)
\( = \sqrt{(3+4)^2+(3+2)^2+(5)^2} \)
\( = \sqrt{7^2+5^2+5^2} \)
\( = \sqrt{49+25+25} \)
\( = \sqrt{99} \)
\( = 3\sqrt{11} \)
In simple words: To find the distance between any two points in 3D space, we subtract their matching coordinates, square each difference, add them up, and then take the square root.

🎯 Exam Tip: Be careful with negative signs when subtracting coordinates, especially when one or both coordinates are negative. A common mistake is not correctly handling `(-a - b)` or `(a - (-b))`. Always remember the 3D distance formula: \( \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2} \).

Question 3. Show that the triangle with vertices (6, 10, 10),(1, 0, -5),(6, -10, 0) is a right-angled triangle, and find its area.
Answer: Let the given vertices of the triangle be A(6, 10, 10), B(1, 0, -5), and C(6, -10, 0). First, we calculate the length of each side using the distance formula.
Length of AB: \( = \sqrt{(1-6)^2+(0-10)^2+(-5-10)^2} \)
\( = \sqrt{(-5)^2+(-10)^2+(-15)^2} \)
\( = \sqrt{25+100+225} \)
\( = \sqrt{350} \)
Length of BC: \( = \sqrt{(6-1)^2+(-10-0)^2+(0-(-5))^2} \)
\( = \sqrt{5^2+(-10)^2+5^2} \)
\( = \sqrt{25+100+25} \)
\( = \sqrt{150} \)
Length of CA: \( = \sqrt{(6-6)^2+(-10-10)^2+(0-10)^2} \)
\( = \sqrt{0^2+(-20)^2+(-10)^2} \)
\( = \sqrt{400+100} \)
\( = \sqrt{500} \)
Now, we check if the Pythagorean theorem holds true for these side lengths. We square each length:
\( AB^2 = 350 \)
\( BC^2 = 150 \)
\( CA^2 = 500 \)
We can see that \( AB^2 + BC^2 = 350 + 150 = 500 \).
Since \( AB^2 + BC^2 = CA^2 \), the triangle ABC is a right-angled triangle, with the right angle at vertex B.
The area of a right-angled triangle is \( \frac{1}{2} \times \text{base} \times \text{height} \). Here, AB and BC are the base and height.
Area of \( \triangle ABC = \frac{1}{2} \times AB \times BC \)
\( = \frac{1}{2} \times \sqrt{350} \times \sqrt{150} \)
\( = \frac{1}{2} \times \sqrt{35 \times 10 \times 15 \times 10} \)
\( = \frac{1}{2} \times \sqrt{35 \times 15 \times 100} \)
\( = \frac{1}{2} \times \sqrt{525 \times 100} \)
\( = \frac{1}{2} \times 10 \sqrt{525} \)
\( = 5 \sqrt{25 \times 21} \)
\( = 5 \times 5 \sqrt{21} \)
\( = 25\sqrt{21} \) square units.
In simple words: First, we find the length of all three sides of the triangle. Then, we check if the square of the longest side is equal to the sum of the squares of the other two sides. If it is, the triangle is right-angled. For the area, we multiply the two shorter sides and divide by two.

🎯 Exam Tip: To prove a triangle is right-angled, calculate the square of each side length. If the sum of the squares of the two shorter sides equals the square of the longest side, then it's a right-angled triangle. The area of a right-angled triangle is half the product of the sides forming the right angle.

Question 4. Show that the triangle with vertices A(3, 5, -4), B(-1, 1, 2), C(-5, -5, -2) is isosceles.
Answer: Let the given vertices of the triangle be A(3, 5, -4), B(-1, 1, 2), and C(-5, -5, -2). A triangle is isosceles if at least two of its sides have equal length. We will calculate the length of each side.
Length of AB: \( = \sqrt{(-1-3)^2+(1-5)^2+(2-(-4))^2} \)
\( = \sqrt{(-4)^2+(-4)^2+(2+4)^2} \)
\( = \sqrt{16+16+36} \)
\( = \sqrt{68} \)
\( = 2\sqrt{17} \)
Length of BC: \( = \sqrt{(-5-(-1))^2+(-5-1)^2+(-2-2)^2} \)
\( = \sqrt{(-5+1)^2+(-6)^2+(-4)^2} \)
\( = \sqrt{(-4)^2+(-6)^2+(-4)^2} \)
\( = \sqrt{16+36+16} \)
\( = \sqrt{68} \)
\( = 2\sqrt{17} \)
Length of CA: \( = \sqrt{(3-(-5))^2+(5-(-5))^2+(-4-(-2))^2} \)
\( = \sqrt{(3+5)^2+(5+5)^2+(-4+2)^2} \)
\( = \sqrt{8^2+10^2+(-2)^2} \)
\( = \sqrt{64+100+4} \)
\( = \sqrt{168} \)
Since AB = BC (both are \( \sqrt{68} \)), the triangle ABC has two sides of equal length. Therefore, triangle ABC is an isosceles triangle.
In simple words: We find the length of each side of the triangle. If any two sides have the same length, then the triangle is an isosceles triangle.

🎯 Exam Tip: To prove a triangle is isosceles, you only need to show that two of its three sides are equal in length. You don't necessarily need to calculate the third side if you find two equal ones first.

Question 5. Show that (4, 2, 4), (10, 2, -2) and (2, 0, -4) are the vertices of an equilateral triangle.
Answer: Let the given vertices of the triangle be A(4, 2, 4), B(10, 2, -2), and C(2, 0, -4). An equilateral triangle has all three sides of equal length. We will calculate the length of each side.
Length of AB: \( = \sqrt{(10-4)^2+(2-2)^2+(-2-4)^2} \)
\( = \sqrt{6^2+0^2+(-6)^2} \)
\( = \sqrt{36+0+36} \)
\( = \sqrt{72} \)
\( = 6\sqrt{2} \)
Length of BC: \( = \sqrt{(2-10)^2+(0-2)^2+(-4-(-2))^2} \)
\( = \sqrt{(-8)^2+(-2)^2+(-4+2)^2} \)
\( = \sqrt{64+4+(-2)^2} \)
\( = \sqrt{64+4+4} \)
\( = \sqrt{72} \)
\( = 6\sqrt{2} \)
Length of CA: \( = \sqrt{(4-2)^2+(2-0)^2+(4-(-4))^2} \)
\( = \sqrt{2^2+2^2+(4+4)^2} \)
\( = \sqrt{4+4+8^2} \)
\( = \sqrt{4+4+64} \)
\( = \sqrt{72} \)
\( = 6\sqrt{2} \)
Since AB = BC = CA (all are \( \sqrt{72} \)), all three sides of the triangle have equal length. Therefore, triangle ABC is an equilateral triangle.
In simple words: We calculate the length of all three sides of the triangle. If all three sides turn out to be the same length, then the triangle is an equilateral triangle.

🎯 Exam Tip: To prove a triangle is equilateral, you must show that all three side lengths are equal. Make sure to clearly write out the distance formula for each side and simplify the square roots.

Question 6. Show that the points (1, -1, 3),(2, -4, 5) and (5, -13, 11) are collinear.
Answer: Let the given points be A(1, -1, 3), B(2, -4, 5), and C(5, -13, 11). Points are collinear if the sum of the distances between two pairs of points is equal to the distance between the remaining pair. We will calculate the distance between each pair of points.
Length of AB: \( = \sqrt{(2-1)^2+(-4-(-1))^2+(5-3)^2} \)
\( = \sqrt{1^2+(-4+1)^2+2^2} \)
\( = \sqrt{1^2+(-3)^2+2^2} \)
\( = \sqrt{1+9+4} \)
\( = \sqrt{14} \)
Length of BC: \( = \sqrt{(5-2)^2+(-13-(-4))^2+(11-5)^2} \)
\( = \sqrt{3^2+(-13+4)^2+6^2} \)
\( = \sqrt{3^2+(-9)^2+6^2} \)
\( = \sqrt{9+81+36} \)
\( = \sqrt{126} \)
\( = \sqrt{9 \times 14} \)
\( = 3\sqrt{14} \)
Length of CA: \( = \sqrt{(1-5)^2+(-1-(-13))^2+(3-11)^2} \)
\( = \sqrt{(-4)^2+(-1+13)^2+(-8)^2} \)
\( = \sqrt{16+12^2+64} \)
\( = \sqrt{16+144+64} \)
\( = \sqrt{224} \)
\( = \sqrt{16 \times 14} \)
\( = 4\sqrt{14} \)
Now, we check if the sum of two smaller distances equals the largest distance:
\( AB + BC = \sqrt{14} + 3\sqrt{14} = 4\sqrt{14} \).
Clearly, \( AB + BC = CA \). This shows that the points A, B, and C lie on a single straight line. Therefore, the points are collinear.
In simple words: We find the distance between every pair of points. If the two shortest distances add up to be the same as the longest distance, then all three points are on the same line.

🎯 Exam Tip: To prove collinearity using distances, always calculate all three distances between the points (AB, BC, CA). Then, check if the sum of the two smaller distances equals the largest distance. If it does, the points are collinear.

Question 7. Derive the equation of the locus of a point equidistant from the points (1, -2, 3) and (-3, 4, 2).
Answer: Let P(x, y, z) be any point on the locus. Let A be the point (1, -2, 3) and B be the point (-3, 4, 2). The condition states that point P is equidistant from A and B, meaning the distance PA is equal to the distance PB.
\( PA = PB \)
We use the distance formula to write this:
\( \sqrt{(x-1)^2+(y-(-2))^2+(z-3)^2} = \sqrt{(x-(-3))^2+(y-4)^2+(z-2)^2} \)
\( \sqrt{(x-1)^2+(y+2)^2+(z-3)^2} = \sqrt{(x+3)^2+(y-4)^2+(z-2)^2} \)
To remove the square roots, we square both sides:
\( (x-1)^2+(y+2)^2+(z-3)^2 = (x+3)^2+(y-4)^2+(z-2)^2 \)
Now, we expand each squared term:
\( (x^2-2x+1) + (y^2+4y+4) + (z^2-6z+9) = (x^2+6x+9) + (y^2-8y+16) + (z^2-4z+4) \)
We can cancel \( x^2, y^2, z^2 \) from both sides:
\( -2x+1+4y+4-6z+9 = 6x+9-8y+16-4z+4 \)
\( -2x+4y-6z+14 = 6x-8y-4z+29 \)
Next, we move all terms to one side to simplify:
\( 0 = 6x+2x-8y-4y-4z+6z+29-14 \)
\( 0 = 8x-12y+2z+15 \)
So, the required equation of the locus of the point is \( 8x-12y+2z+15 = 0 \). This equation represents a plane that is the perpendicular bisector of the line segment AB.
In simple words: A point that is always the same distance from two other points will form a flat surface (a plane). We find the equation for this surface by setting the distance from our moving point to the first fixed point equal to its distance to the second fixed point, and then we simplify the equation.

🎯 Exam Tip: When deriving a locus equation, always set up the given condition (e.g., equidistant, distance ratio) as an equation. Squaring both sides is often the first step to eliminate square roots, then expand and simplify carefully, canceling terms common to both sides.

Question 8. Derive the equation of the locus of a point twice as far from (-2, 3, 4) as from (3, -1, -2).
Answer: Let P(x, y, z) be any point on the locus. Let A be the point (-2, 3, 4) and B be the point (3, -1, -2). The condition states that point P is twice as far from A as it is from B.
\( PA = 2PB \)
We write this using the distance formula:
\( \sqrt{(x-(-2))^2+(y-3)^2+(z-4)^2} = 2 \sqrt{(x-3)^2+(y-(-1))^2+(z-(-2))^2} \)
\( \sqrt{(x+2)^2+(y-3)^2+(z-4)^2} = 2 \sqrt{(x-3)^2+(y+1)^2+(z+2)^2} \)
To eliminate the square roots, we square both sides:
\( (x+2)^2+(y-3)^2+(z-4)^2 = 4[(x-3)^2+(y+1)^2+(z+2)^2] \)
Now, we expand each squared term:
\( (x^2+4x+4) + (y^2-6y+9) + (z^2-8z+16) = 4[(x^2-6x+9) + (y^2+2y+1) + (z^2+4z+4)] \)
\( x^2+y^2+z^2+4x-6y-8z+29 = 4[x^2+y^2+z^2-6x+2y+4z+14] \)
\( x^2+y^2+z^2+4x-6y-8z+29 = 4x^2+4y^2+4z^2-24x+8y+16z+56 \)
Move all terms to one side, typically to the side where the squared terms remain positive:
\( 0 = (4x^2-x^2) + (4y^2-y^2) + (4z^2-z^2) + (-24x-4x) + (8y-(-6y)) + (16z-(-8z)) + (56-29) \)
\( 0 = 3x^2+3y^2+3z^2-28x+14y+24z+27 \)
Thus, the required equation of the locus is \( 3x^2+3y^2+3z^2-28x+14y+24z+27 = 0 \). This equation represents a sphere.
In simple words: If a point is always twice as far from one fixed point as it is from another fixed point, its path forms a special curved surface. We find the equation for this surface by using the distance formula and making one distance double the other, then simplifying.

🎯 Exam Tip: When the distance ratio is not 1:1, remember to square the ratio (e.g., \( (2)^2 = 4 \)) when squaring both sides of the distance equation. Be meticulous with algebraic expansion and simplification to avoid errors in the final quadratic equation.

Question 9. Find the equation of the locus of a point whose distance from the y-axis is equal to its distance from (2, 1, -1).
Answer: Let P(x, y, z) be any point on the locus. Let Q(0, y, 0) be the point on the y-axis closest to P (this is the foot of the perpendicular from P to the y-axis). Let A be the given point (2, 1, -1). The condition states that the distance from P to the y-axis (PQ) is equal to the distance from P to A (PA).
\( PQ = PA \)
Using the distance formula:
\( \sqrt{(x-0)^2+(y-y)^2+(z-0)^2} = \sqrt{(x-2)^2+(y-1)^2+(z-(-1))^2} \)
\( \sqrt{x^2+0^2+z^2} = \sqrt{(x-2)^2+(y-1)^2+(z+1)^2} \)
\( \sqrt{x^2+z^2} = \sqrt{(x-2)^2+(y-1)^2+(z+1)^2} \)
To remove the square roots, we square both sides:
\( x^2+z^2 = (x-2)^2+(y-1)^2+(z+1)^2 \)
Now, we expand the squared terms on the right side:
\( x^2+z^2 = (x^2-4x+4) + (y^2-2y+1) + (z^2+2z+1) \)
\( x^2+z^2 = x^2+y^2+z^2-4x-2y+2z+6 \)
We can cancel \( x^2 \) and \( z^2 \) from both sides:
\( 0 = y^2-4x-2y+2z+6 \)
So, the required equation of the locus is \( y^2-4x-2y+2z+6 = 0 \). This equation describes a parabolic cylinder.
In simple words: Imagine a point moving in space. If its shortest distance to the y-axis is always the same as its distance to a specific fixed point, then its path creates a curved surface. We find the math rule for this surface by setting those distances equal and simplifying.

🎯 Exam Tip: The distance of a point (x, y, z) from the x-axis is \( \sqrt{y^2+z^2} \), from the y-axis is \( \sqrt{x^2+z^2} \), and from the z-axis is \( \sqrt{x^2+y^2} \). Remember these specific distance formulas for axes to correctly set up the initial equation.

Question 10. Find the equation of the locus of a point whose distance from the xy-plane is equal to distance from the point (-1, 2, -3).
Answer: Let P(x, y, z) be any point on the locus. Let Q(x, y, 0) be the point on the xy-plane closest to P (this is the foot of the perpendicular from P to the xy-plane). Let B be the given point (-1, 2, -3). The condition states that the distance from P to the xy-plane (PQ) is equal to the distance from P to B (PB).
\( PQ = PB \)
Using the distance formula:
\( \sqrt{(x-x)^2+(y-y)^2+(z-0)^2} = \sqrt{(x-(-1))^2+(y-2)^2+(z-(-3))^2} \)
\( \sqrt{0^2+0^2+z^2} = \sqrt{(x+1)^2+(y-2)^2+(z+3)^2} \)
\( \sqrt{z^2} = \sqrt{(x+1)^2+(y-2)^2+(z+3)^2} \)
So, \( |z| = \sqrt{(x+1)^2+(y-2)^2+(z+3)^2} \). To remove the square root, we square both sides:
\( z^2 = (x+1)^2+(y-2)^2+(z+3)^2 \)
Now, we expand the squared terms on the right side:
\( z^2 = (x^2+2x+1) + (y^2-4y+4) + (z^2+6z+9) \)
\( z^2 = x^2+y^2+z^2+2x-4y+6z+14 \)
We can cancel \( z^2 \) from both sides:
\( 0 = x^2+y^2+2x-4y+6z+14 \)
Thus, the required equation of the locus is \( x^2+y^2+2x-4y+6z+14 = 0 \). This equation represents a paraboloid.
In simple words: When a point moves so that its distance to the flat xy-plane is always the same as its distance to a specific point, its path forms a curved shape. We find the equation for this shape by setting those two distances equal and simplifying.

🎯 Exam Tip: The distance of a point (x, y, z) from the xy-plane is \( |z| \), from the yz-plane is \( |x| \), and from the xz-plane is \( |y| \). Use this simple fact to quickly set up the left side of the equation when dealing with distances to coordinate planes.

Question 11. A point moves so that the difference of the squares of its distances from the x-axis and the y-axis is constant. Find the equation of its locus.
Answer: Let P(x, y, z) be any point on the locus. Let Q(x, 0, 0) be the point on the x-axis closest to P, and R(0, y, 0) be the point on the y-axis closest to P. The condition states that the difference of the squares of its distances from the x-axis (PQ) and the y-axis (PR) is a constant, k.
\( PQ^2 - PR^2 = k \)
First, we find the distance PQ (distance from P to x-axis):
\( PQ = \sqrt{(x-x)^2+(y-0)^2+(z-0)^2} = \sqrt{0^2+y^2+z^2} = \sqrt{y^2+z^2} \)
So, \( PQ^2 = y^2+z^2 \).
Next, we find the distance PR (distance from P to y-axis):
\( PR = \sqrt{(x-0)^2+(y-y)^2+(z-0)^2} = \sqrt{x^2+0^2+z^2} = \sqrt{x^2+z^2} \)
So, \( PR^2 = x^2+z^2 \).
Now, we substitute these into the given condition:
\( (y^2+z^2) - (x^2+z^2) = k \)
Remove the parentheses:
\( y^2+z^2-x^2-z^2 = k \)
The \( z^2 \) terms cancel each other out:
\( y^2-x^2 = k \)
Thus, the required equation of the locus is \( y^2-x^2 = k \). This equation represents a hyperbolic cylinder.
In simple words: Imagine a point moving in a way that if you square its distance to the x-axis, then square its distance to the y-axis, and subtract the second from the first, you always get the same number. The path this point traces out is called its locus, and its equation is simply \( y^2-x^2 = k \).

🎯 Exam Tip: When dealing with squares of distances, remember that \( (\sqrt{a})^2 = a \), so the square roots cancel out directly. Be careful with signs when subtracting algebraic expressions, ensuring all terms in the subtracted expression have their signs flipped.

Question 12. Find the equation of the locus of a point whose distance from the z-axis is equal to its distance from the xy-plane.
Answer: Let P(x, y, z) be any point on the locus. Let Q(0, 0, z) be the point on the z-axis closest to P. Let R(x, y, 0) be the point on the xy-plane closest to P. The condition states that the distance from P to the z-axis (PQ) is equal to the distance from P to the xy-plane (PR).
\( PQ = PR \)
First, we find PQ (distance from P to z-axis):
\( PQ = \sqrt{(x-0)^2+(y-0)^2+(z-z)^2} = \sqrt{x^2+y^2+0^2} = \sqrt{x^2+y^2} \)
Next, we find PR (distance from P to xy-plane):
\( PR = \sqrt{(x-x)^2+(y-y)^2+(z-0)^2} = \sqrt{0^2+0^2+z^2} = \sqrt{z^2} = |z| \)
Now, we set these two distances equal:
\( \sqrt{x^2+y^2} = |z| \)
To remove the square root and absolute value, we square both sides:
\( (\sqrt{x^2+y^2})^2 = (|z|)^2 \)
\( x^2+y^2 = z^2 \)
We can rearrange this equation:
\( x^2+y^2-z^2 = 0 \)
Thus, the required equation of the locus is \( x^2+y^2-z^2 = 0 \). This equation represents a double cone with its vertex at the origin.
In simple words: If a point moves so that its shortest distance to the z-axis is always the same as its shortest distance to the flat xy-plane, then its path forms a cone shape. We find the equation for this cone by setting those distances equal and simplifying.

🎯 Exam Tip: Always correctly identify the point on the axis or plane that gives the shortest distance. For the z-axis, it's (0,0,z). For the xy-plane, it's (x,y,0). This is crucial for setting up the distance formulas correctly.