CBSE Class 11 Mathematics Three Dimensional Geometry Assignment Set 01

Question 1. (i) If \((x-y, 3) = (5, x+y)\) then find the value of \(x\) and \(y\).
Answer: Comparing the corresponding coordinates of the equal ordered pairs:
\(x - y = 5\)
\(x + y = 3\)
Adding the two equations: \(2x = 8 \implies x = 4\)
Substituting \(x = 4\) in \(x + y = 3\): \(4 + y = 3 \implies y = -1\)
Thus, \(x = 4\) and \(y = -1\).
In simple words: Equating corresponding components gives a linear system of two variables.
Exam Tip: Two ordered pairs \((a, b)\) and \((c, d)\) are equal if and only if \(a = c\) and \(b = d\).

Question 1. (ii) If \((x+3, 5) = (6, 2x+y)\) then find the value of \(x\) and \(y\).
Answer: Equating corresponding components:
\(x + 3 = 6 \implies x = 3\)
\(2x + y = 5\)
Substituting \(x = 3\) into the second equation: \(2(3) + y = 5 \implies 6 + y = 5 \implies y = -1\)
Thus, \(x = 3\) and \(y = -1\).
In simple words: Solve for \(x\) first, then substitute its value to find \(y\).

Exam Tip: Always solve for the single variable equation first to save time.

Question 1. (iii) If \((x-1, y+3) = (2, -5)\) then find the value of \(x\) and \(y\).
Answer: Equating corresponding components:
\(x - 1 = 2 \implies x = 3\)
\(y + 3 = -5 \implies y = -8\)
Thus, \(x = 3\) and \(y = -8\).
In simple words: Each coordinate can be solved independently of the other.

Exam Tip: Verify by substituting back into the original ordered pair.

Question 2. If \(n(A) = p\), \(n(B) = q\), \(n(C) = r\) and \(n(D) = s\) then find (i) \(n(A \times B \times C \times D)\) (ii) Number of possible subsets of \(A \times B \times C \times D\) (iii) \(n[P(A \times B \times C \times D)]\) (iv) Number of possible relations from \(A\) to \(C\) (v) Number of possible relations from \(B\) to \(D\)
Answer:
(i) \(n(A \times B \times C \times D) = n(A) \cdot n(B) \cdot n(C) \cdot n(D) = pqrs\)
(ii) Number of possible subsets of a set with \(n\) elements is \(2^n\). Here, number of elements is \(pqrs\), so number of subsets = \(2^{pqrs}\).
(iii) The power set \(P(S)\) is the set of all subsets of \(S\). Thus, \(n[P(A \times B \times C \times D)] = 2^{pqrs}\).
(iv) Number of relations from \(A\) to \(C\) is equal to the number of subsets of \(A \times C\), which is \(2^{n(A) \cdot n(C)} = 2^{pr}\).
(v) Number of relations from \(B\) to \(D\) is equal to the number of subsets of \(B \times D\), which is \(2^{n(B) \cdot n(D)} = 2^{qs}\).
In simple words: Use cardinalities and exponential formulas for Cartesian products and power sets.

Exam Tip: Remember that the number of relations from set \(X\) to set \(Y\) is \(2^{n(X) \cdot n(Y)}\).

Question 3. If \(A = \{1, 2, 3\}\), \(B = \{3, 4\}\) and \(C = \{1, 3, 5\}\), find (i) \(A \times (B \cup C)\) (ii) \(A \times (B \cap C)\) (iii) \((A \times B) \cap (B \times C)\)
Answer:
First, find set operations on \(B\) and \(C\):
\(B \cup C = \{1, 3, 4, 5\}\)
\(B \cap C = \{3\}\)
(i) \(A \times (B \cup C) = \{1, 2, 3\} \times \{1, 3, 4, 5\} = \{(1,1), (1,3), (1,4), (1,5), (2,1), (2,3), (2,4), (2,5), (3,1), (3,3), (3,4), (3,5)\}\)
(ii) \(A \times (B \cap C) = \{1, 2, 3\} \times \{3\} = \{(1,3), (2,3), (3,3)\}\)
(iii) First find \(A \times B = \{(1,3), (1,4), (2,3), (2,4), (3,3), (3,4)\}\) and \(B \times C = \{(3,1), (3,3), (3,5), (4,1), (4,3), (4,5)\}\).
Their intersection is the set of common ordered pairs: \((A \times B) \cap (B \times C) = \{(3,3)\}\).
In simple words: Perform set union and intersection first, then compute the Cartesian products.

Exam Tip: Distributive laws like \(A \times (B \cap C) = (A \times B) \cap (A \times C)\) can also be used as alternative methods.

Question 4. A relation \(R\) is defined from a set \(A = \{2, 3, 4, 5\}\) to a set \(B = \{3, 6, 7, 10\}\) as follows \(\{(x, y) : x \text{ divides } y\}\). Express \(R\) as a set of ordered pairs and determine the domain and range of \(R\).
Answer:
Check each element of \(A\) against each element of \(B\) to see if \(x\) divides \(y\) (\(y\) is a multiple of \(x\)):
- For \(2 \in A\): divides \(6\) and \(10 \implies (2,6), (2,10)\)
- For \(3 \in A\): divides \(3\) and \(6 \implies (3,3), (3,6)\)
- For \(4 \in A\): does not divide any element in \(B\)
- For \(5 \in A\): divides \(10 \implies (5,10)\)
Thus, \(R = \{(2,6), (2,10), (3,3), (3,6), (5,10)\}\).
Domain = Set of all first elements = \(\{2, 3, 5\}\)
Range = Set of all second elements = \(\{3, 6, 10\}\)
In simple words: Find pairs where the first number evenly divides the second number.

Exam Tip: Domain is a subset of \(A\) and range is a subset of \(B\).

Question 5. Find the domain and range of following functions:-
(i) \(f(x) = \sqrt{x^2 - 4}\)
Answer:
For domain: \(x^2 - 4 \geq 0 \implies (x-2)(x+2) \geq 0 \implies x \in (-\infty, -2] \cup [2, \infty)\).
For range: As \(x^2 - 4\) varies from \(0\) to \(\infty\), its square root varies from \(0\) to \(\infty\). Thus, Range = \([0, \infty)\).
In simple words: Expression under root must be non-negative.

Exam Tip: Check endpoints of the inequality for domain boundaries.

(ii) \(f(x) = \frac{1}{\sqrt{9 - x^2}}\)
Answer:
For domain: \(9 - x^2 > 0 \implies x^2 < 9 \implies x \in (-3, 3)\).
For range: As \(x\) goes from \(-3\) to \(3\), \(9 - x^2\) goes from \(0\) to \(9\), so \(\sqrt{9 - x^2}\) goes from \(0\) to \(3\). Hence, the reciprocal goes from \(\frac{1}{3}\) to \(\infty\). Range = \([\frac{1}{3}, \infty)\).
In simple words: Denominator cannot be zero, and radicand must be strictly positive.

Exam Tip: Denominator strictly greater than zero because of the square root in the denominator.

(iii) \(f(x) = \sqrt{1 - x^2}\)
Answer:
For domain: \(1 - x^2 \geq 0 \implies x^2 \leq 1 \implies x \in [-1, 1]\).
For range: Maximum value is at \(x = 0\) which gives \(\sqrt{1} = 1\), and minimum value is at endpoints \(x = \pm 1\) which gives \(0\). Range = \([0, 1]\).
In simple words: Represents the upper half of a unit circle.

Exam Tip: Sine and cosine substitutions often simplify radical functions.

(iv) \(f(x) = 4\sin x - 3\cos x\)
Answer:
For domain: Trigonometric functions are defined for all real numbers, so Domain = Real Numbers (\(\mathbb{R}\)).
For range: Any function of the form \(a\sin x + b\cos x\) has range \([-\sqrt{a^2 + b^2}, \sqrt{a^2 + b^2}]\).
Here, \(\sqrt{4^2 + (-3)^2} = \sqrt{16 + 9} = \sqrt{25} = 5\). Range = \([-5, 5]\).
In simple words: Use the standard amplitude formula \(\sqrt{a^2 + b^2}\) for sine-cosine linear combinations.

Exam Tip: Memorize the auxiliary angle method and range bounds for trigonometric expressions.

(v) \(f(x) = 1 + 3\cos(2x)\)
Answer:
For domain: Defined for all real numbers, Domain = Real Numbers (\(\mathbb{R}\)).
For range: Since \(-1 \leq \cos(2x) \leq 1\), multiplying by \(3\) gives \(-3 \leq 3\cos(2x) \leq 3\). Adding \(1\) gives \(-2 \leq 1 + 3\cos(2x) \leq 4\). Range = \([-2, 4]\).
In simple words: Scale and shift the known bounds of the cosine function.

Exam Tip: Range of cosine is always \([-1, 1]\) regardless of the angle multiplier.

(vi) \(f(x) = |x - 3|\)
Answer:
For domain: Absolute value is defined everywhere, Domain = Real Numbers (\(\mathbb{R}\)).
For range: Absolute value output is always non-negative. Range = \([0, \infty)\).
In simple words: Modulus function output cannot be negative.

Exam Tip: The graph of \(|x - 3|\) is a V-shape with vertex at \((3, 0)\).

(vii) \(f(x) = \frac{x^2}{x^2 + 1}\)
Answer:
For domain: Since \(x^2 + 1 \geq 1 > 0\) for all real \(x\), the denominator is never zero. Domain = Real Numbers (\(\mathbb{R}\)).
For range: Let \(y = \frac{x^2}{x^2 + 1} \implies yx^2 + y = x^2 \implies x^2(1 - y) = y \implies x^2 = \frac{y}{1 - y}\).
Since \(x^2 \geq 0\), we must have \(\frac{y}{1 - y} \geq 0 \implies y \in [0, 1)\). Range = \([0, 1)\).
In simple words: Express \(x^2\) in terms of \(y\) and set it greater than or equal to zero.

Exam Tip: Notice that numerator is strictly less than denominator for all real \(x\), so \(y < 1\).

(viii) \(f(x) = 3x + 7\)
Answer:
For domain: Linear polynomial is defined for all real numbers, Domain = Real Numbers (\(\mathbb{R}\)).
For range: As \(x\) takes all real values, \(3x + 7\) also takes all real values. Range = Real Numbers (\(\mathbb{R}\)).
In simple words: Linear functions with non-zero slope map reals to reals.

Exam Tip: All non-constant linear functions have domain and range as \(\mathbb{R}\).

(ix) \(f(x) = x^2 + 1\)
Answer:
For domain: Polynomial function defined everywhere, Domain = Real Numbers (\(\mathbb{R}\)).
For range: Since \(x^2 \geq 0\) for all real \(x\), adding \(1\) gives \(x^2 + 1 \geq 1\). Range = \([1, \infty)\).
In simple words: Minimum value occurs at \(x = 0\).

Exam Tip: Parabola opens upwards with vertex at \((0, 1)\).

(x) \(f(x) = \frac{|x-4|}{x-4}\)
Answer:
For domain: Denominator \(x - 4 \neq 0 \implies x \neq 4\). Domain = \((-\infty, 4) \cup (4, \infty)\) or \(\mathbb{R} - \{4\}\).
For range: When \(x > 4\), \(|x-4| = x-4\), so \(f(x) = 1\). When \(x < 4\), \(|x-4| = -(x-4)\), so \(f(x) = -1\). Range = \(\{-1, 1\}\).
In simple words: This is the signum-like function for \((x-4)\), yielding only \(1\) or \(-1\).

Exam Tip: Always check points around the critical value where the absolute value expression changes sign.

(xi) \(f(x) = -|x|\)
Answer:
For domain: Defined for all real numbers, Domain = Real Numbers (\(\mathbb{R}\)).
For range: Since \(|x| \geq 0\), multiplying by \(-1\) gives \(-|x| \leq 0\). Range = \((-\infty, 0]\).
In simple words: Inverts the standard absolute value graph to open downwards.

(xii) \(f(x) = \frac{x-2}{x+1}\)
Answer:
For domain: Denominator \(x + 1 \neq 0 \implies x \neq -1\). Domain = \(\mathbb{R} - \{-1\}\) or \((-\infty, -1) \cup (-1, \infty)\).
For range: Let \(y = \frac{x-2}{x+1} \implies y(x+1) = x-2 \implies yx + y = x - 2 \implies x(1-y) = y+2 \implies x = \frac{y+2}{1-y}\).
For \(x\) to be real, \(1 - y \neq 0 \implies y \neq 1\). Range = \(\mathbb{R} - \{1\}\) or \((-\infty, 1) \cup (1, \infty)\).
In simple words: Solve for \(x\) in terms of \(y\) to find restrictions on the output variable.

Exam Tip: For rational functions of degree \(1/1\), range excludes the ratio of leading coefficients (\(1/1 = 1\)).

Question 6. If \(A = \{-2, -1, 1, 2\}\) and \(f = \{(x, \frac{1}{x}) : x \in A\}\), find domain and range of \(f\).
Answer:
First, construct the function values for each \(x \in A\):
- For \(x = -2\), \(f(-2) = -\frac{1}{2}\)
- For \(x = -1\), \(f(-1) = -1\)
- For \(x = 1\), \(f(1) = 1\)
- For \(x = 2\), \(f(2) = \frac{1}{2}\)
Thus, \(f = \{(-2, -\frac{1}{2}), (-1, -1), (1, 1), (2, \frac{1}{2})\}\).
Domain = Set of first elements = \(\{-2, -1, 1, 2\} = A\).
Range = Set of second elements = \(\{-\frac{1}{2}, -1, 1, \frac{1}{2}\}\) or \(\{-\frac{1}{2}, -1, 1, \frac{1}{2}\}\).
In simple words: Domain is the given set \(A\), and range is the set of reciprocals of elements in \(A\).

Exam Tip: Always list domain and range as sets in roster or builder form.

Question 7. Find the domain for which the function \(f(x) = 3x^2 - 1\) and \(g(x) = 3 + x\) are equal.
Answer:
Set \(f(x) = g(x)\):
\(3x^2 - 1 = 3 + x\)
\(3x^2 - x - 4 = 0\)
Factorize the quadratic equation:
\(3x^2 - 4x + 3x - 4 = 0\)
\(x(3x - 4) + 1(3x - 4) = 0\)
\((3x - 4)(x + 1) = 0\)
Thus, \(x = \frac{4}{3}\) or \(x = -1\).
Domain = \(\{-1, \frac{4}{3}\}\).
In simple words: Equate the two function expressions and solve the resulting quadratic equation.

Exam Tip: Verify both solutions by plugging them back into both original functions.

Question 8. Let \(f(x) = 3x + 4 \ \forall x \in \mathbb{R}\). If ordered pairs \((a, 8)\) and \((2, b)\) belong to \(f\). Find \(a\) and \(b\).
Answer:
Since \((a, 8) \in f\), we have \(f(a) = 8\):
\(3a + 4 = 8 \implies 3a = 4 \implies a = \frac{4}{3}\).
Since \((2, b) \in f\), we have \(f(2) = b\):
\(b = 3(2) + 4 = 6 + 4 = 10\).
Thus, \(a = \frac{4}{3}\) and \(b = 10\).
In simple words: Substitute the given \(x\) and \(y\) coordinates into the function rule.

Exam Tip: An ordered pair \((x, y)\) belonging to function \(f\) means \(f(x) = y\).

Question 9. Draw the graph of each of following functions is given by:-
(i) \(f(x) = \frac{1}{x}, x \neq 0\)
Answer:
This is a rectangular hyperbola with branches in the first and third quadrants. It has vertical asymptote \(x = 0\) and horizontal asymptote \(y = 0\).
In simple words: Plot points like \((1,1), (2,0.5), (-1,-1)\) and draw smooth curves approaching axes.

Exam Tip: Note that the graph never touches either coordinate axis.

(ii) \(f(x) = \frac{x}{|x|}\)
Answer:
For \(x > 0\), \(f(x) = 1\). For \(x < 0\), \(f(x) = -1\). At \(x = 0\), the function is undefined.
In simple words: Two horizontal ray segments at \(y = 1\) and \(y = -1\) with a hole at \(x = 0\).

Exam Tip: This is the signum function without the origin point \((0,0)\).

(iii) \(f(x) = |x - 2| + |x - 3|\)
Answer:
Break into intervals based on critical points \(x = 2\) and \(x = 3\):
- For \(x < 2\): \(f(x) = -(x-2) - (x-3) = -2x + 5\)
- For \(2 \leq x \leq 3\): \(f(x) = (x-2) - (x-3) = 1\)
- For \(x > 3\): \(f(x) = (x-2) + (x-3) = 2x - 5\)
In simple words: The graph forms a trough (bathtub shape) with a flat bottom of height \(1\) between \(2\) and \(3\).

Exam Tip: Analyze the slopes in each interval to easily sketch piecewise linear graphs.

(iv) \(f(x) = \sqrt{9 - x^2}\)
Answer:
Squaring both sides gives \(y^2 = 9 - x^2 \implies x^2 + y^2 = 9\) with the restriction \(y \geq 0\). This is the upper semicircle of radius \(3\) centered at the origin.
In simple words: Draw a semicircle above the x-axis connecting \((-3,0)\) to \((3,0)\) through \((0,3)\).

Exam Tip: Recognize circle equations in standard form \(x^2 + y^2 = r^2\).

(v) \(f(x) = x - [x]\)
Answer:
This is the fractional part function. It is periodic with period \(1\). On any interval \([n, n+1)\), \(f(x) = x - n\).
In simple words: A repeating sawtooth wave pattern between \(0\) (inclusive) and \(1\) (exclusive).

Exam Tip: Use open circles at the upper end of each repeating segment to show strict inequality.

(vi) \(f(x) = |x|\)
Answer:
For \(x \geq 0\), \(f(x) = x\). For \(x < 0\), \(f(x) = -x\).
In simple words: A V-shaped graph with vertex at the origin \((0,0)\) making \(45^\circ\) angles with axes.

Exam Tip: The absolute value function is symmetric about the y-axis (even function).

Question 10. If \(A = \{a, b, c\}\), \(B = \{d\}\) and \(C = \{e\}\). Verify the following:-
(i) \(A \times (B \cup C) = (A \times B) \cup (A \times C)\)
Answer:
- Left Hand Side (LHS):
\(B \cup C = \{d, e\}\)
\(A \times (B \cup C) = \{a, b, c\} \times \{d, e\} = \{(a,d), (a,e), (b,d), (b,e), (c,d), (c,e)\}\)
- Right Hand Side (RHS):
\(A \times B = \{(a,d), (b,d), (c,d)\}\)
\(A \times C = \{(a,e), (b,e), (c,e)\}\)
\((A \times B) \cup (A \times C) = \{(a,d), (b,d), (c,d), (a,e), (b,e), (c,e)\}\)
Since LHS = RHS, the distributive property is verified.
In simple words: Cartesian product distributes over set union.

Exam Tip: Set properties mirror algebraic distributive laws.

(ii) \((A \cup B) \times C = (A \times C) \cup (B \times C)\)
Answer:
- Left Hand Side (LHS):
\(A \cup B = \{a, b, c, d\}\)
\((A \cup B) \times C = \{a, b, c, d\} \times \{e\} = \{(a,e), (b,e), (c,e), (d,e)\}\)
- Right Hand Side (RHS):
\(A \times C = \{(a,e), (b,e), (c,e)\}\)
\(B \times C = \{(d,e)\}\)
\((A \times C) \cup (B \times C) = \{(a,e), (b,e), (c,e), (d,e)\}\)
Since LHS = RHS, the property is verified.
In simple words: Cartesian product also distributes over union from the left.

Exam Tip: Always compute set operations inside parentheses first.

Question 11. Determine the domain and the range of the relation \(R\) defined by \(R = \{(x+1, x+5) : x \in \{0, 1, 2, 3, 4, 5\}\}\).
Answer:
Substitute each value of \(x \in \{0, 1, 2, 3, 4, 5\}\) into the ordered pair \((x+1, x+5)\):
- For \(x = 0\): \((1, 5)\)
- For \(x = 1\): \((2, 6)\)
- For \(x = 2\): \((3, 7)\)
- For \(x = 3\): \((4, 8)\)
- For \(x = 4\): \((5, 9)\)
- For \(x = 5\): \((6, 10)\)
Thus, \(R = \{(1,5), (2,6), (3,7), (4,8), (5,9), (6,10)\}\).
Domain = Set of first components = \(\{1, 2, 3, 4, 5, 6\}\).
Range = Set of second components = \(\{5, 6, 7, 8, 9, 10\}\).
In simple words: Generate all ordered pairs by plugging in given \(x\) values, then collect first and second elements.

Exam Tip: Notice that each domain element \(x+1\) corresponds directly to range element \(x+5\).

Question 12. Is \(g = \{(1,1), (2,3), (3,5), (4,7)\}\) a function? If this is described by the formula \(g(x) = ax + b\) then what values should be assigned to \(a\) and \(b\)?
Answer:
Yes, \(g\) is a function because every element in the domain \(\{1, 2, 3, 4\}\) has a unique image in the range.
Using the formula \(g(x) = ax + b\):
Using the pair \((1, 1)\): \(a(1) + b = 1 \implies a + b = 1\) -- (Equation 1)
Using the pair \((2, 3)\): \(a(2) + b = 3 \implies 2a + b = 3\) -- (Equation 2)
Subtracting Equation 1 from Equation 2:
\((2a + b) - (a + b) = 3 - 1 \implies a = 2\).
Substitute \(a = 2\) into Equation 1:
\(2 + b = 1 \implies b = -1\).
Thus, \(a = 2\) and \(b = -1\).
In simple words: Check uniqueness of first elements for function definition, then set up simultaneous linear equations for \(a\) and \(b\).

Question 13. If \(R\) is the relation 'is greater than' from \(A = \{2, 3, 4, 5, 6\}\) to \(B = \{2, 5, 6\}\). Write \(R\) in tabular form, its domain and range.
Answer:
Find pairs \((x, y) \in A \times B\) such that \(x > y\):
- For \(x = 2\): none in \(B\) is less than \(2\).
- For \(x = 3\): \(3 > 2 \implies (3, 2)\).
- For \(x = 4\): \(4 > 2 \implies (4, 2)\).
- For \(x = 5\): \(5 > 2 \implies (5, 2)\).
- For \(x = 6\): \(6 > 2\) and \(6 > 5 \implies (6, 2), (6, 5)\).
Thus, \(R = \{(3,2), (4,2), (5,2), (6,2), (6,5)\}\).
Domain = \(\{3, 4, 5, 6\}\).
Range = \(\{2, 5\}\).
In simple words: List pairs where the first element is strictly greater than the second element.

Exam Tip: Tabular form means listing the relation as a set of ordered pairs.

Question 14. Given \(A = \{-1, 0, 2, 5, 6, 11\}\), \(B = \{-2, -1, 0, 18, 28, 108\}\) and \(f(x) = x^2 - x - 2\). Find \(f(A)\). Is \(f(A) = B\)?
Answer:
Evaluate \(f(x) = x^2 - x - 2\) for each element in \(A\):
- \(f(-1) = (-1)^2 - (-1) - 2 = 1 + 1 - 2 = 0\)
- \(f(0) = (0)^2 - 0 - 2 = -2\)
- \(f(2) = (2)^2 - 2 - 2 = 4 - 4 = 0\)
- \(f(5) = (5)^2 - 5 - 2 = 25 - 7 = 18\)
- \(f(6) = (6)^2 - 6 - 2 = 36 - 8 = 28\)
- \(f(11) = (11)^2 - 11 - 2 = 121 - 13 = 108\)
Thus, \(f(A) = \{0, -2, 0, 18, 28, 108\} = \{-2, 0, 18, 28, 108\}\) (removing duplicate \(0\)).
Comparing with \(B = \{-2, -1, 0, 18, 28, 108\}\): Since \(-1 \in B\) but \(-1 \notin f(A)\), \(f(A) \neq B\).
In simple words: Calculate function outputs for all elements in set \(A\) and check if the resulting set equals set \(B\).

Exam Tip: Set equality requires every element of one set to be present in the other.

Question 15. Show that the points \((-2, 6, -2)\), \((0, 4, -1)\), \((-2, 3, 1)\) and \((-4, 5, 0)\) are vertices of the square.
Answer:
Let \(A(-2, 6, -2)\), \(B(0, 4, -1)\), \(C(-2, 3, 1)\), and \(D(-4, 5, 0)\).
Calculate side lengths using the distance formula \(d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}\):
- \(AB = \sqrt{(0 - (-2))^2 + (4 - 6)^2 + (-1 - (-2))^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3\)
- \(BC = \sqrt{(-2 - 0)^2 + (3 - 4)^2 + (1 - (-1))^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3\)
- \(CD = \sqrt{(-4 - (-2))^2 + (5 - 3)^2 + (0 - 1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3\)
- \(DA = \sqrt{(-2 - (-4))^2 + (6 - 5)^2 + (-2 - 0)^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3\)
All four sides are equal: \(AB = BC = CD = DA = 3\).
Now check the diagonals:
- \(AC = \sqrt{(-2 - (-2))^2 + (3 - 6)^2 + (1 - (-2))^2} = \sqrt{0 + 9 + 9} = \sqrt{18} = 3\sqrt{2}\)
- \(BD = \sqrt{(-4 - 0)^2 + (5 - 4)^2 + (0 - (-1))^2} = \sqrt{16 + 1 + 1} = \sqrt{18} = 3\sqrt{2}\)
Since all four sides are equal and both diagonals are equal, the points form a square.
In simple words: A quadrilateral with four equal sides and equal diagonals is a square.

Exam Tip: Always check both sides and diagonals to distinguish a square from a rhombus.

Question 16. Find the ratio in which the line segment joining the points \((2, 4, -3)\) and \((-3, 5, 4)\) is divided by (i) the \(xy\)-plane (ii) the plane \(x + y + z = 8\).
Answer:
Let the ratio be \(k : 1\). The coordinates of the dividing point are given by the section formula:
\(\left( \frac{-3k + 2}{k+1}, \frac{5k + 4}{k+1}, \frac{4k - 3}{k+1} \right)\)
(i) For the \(xy\)-plane, the \(z\)-coordinate is \(0\):
\(\frac{4k - 3}{k+1} = 0 \implies 4k - 3 = 0 \implies k = \frac{3}{4}\).
Ratio = \(\frac{3}{4} : 1 = 3 : 4\) internally.
(ii) For the plane \(x + y + z = 8\), substitute coordinates into the plane equation:
\(\frac{-3k + 2}{k+1} + \frac{5k + 4}{k+1} + \frac{4k - 3}{k+1} = 8\)
\(\frac{-3k + 2 + 5k + 4 + 4k - 3}{k+1} = 8\)
\(\frac{6k + 3}{k+1} = 8\)
\(6k + 3 = 8k + 8 \implies 2k = -5 \implies k = -\frac{5}{2}\).
Ratio = \(5 : 2\) externally (since \(k\) is negative).
In simple words: Set \(z = 0\) for the \(xy\)-plane, and substitute general section coordinates into the plane equation for the second part.

Exam Tip: A positive ratio indicates internal division, while a negative ratio indicates external division.

Question 17. Show that the three points \(A(1, -2, -8)\), \(B(5, 0, -2)\) and \(C(11, 3, 7)\) are collinear. Find the ratio in which \(B\) divides \(AC\).
Answer:
Calculate distances between the points:
- \(AB = \sqrt{(5-1)^2 + (0 - (-2))^2 + (-2 - (-8))^2} = \sqrt{16 + 4 + 36} = \sqrt{56} = 2\sqrt{14}\)
- \(BC = \sqrt{(11-5)^2 + (3-0)^2 + (7 - (-2))^2} = \sqrt{36 + 9 + 81} = \sqrt{126} = 3\sqrt{14}\)
- \(AC = \sqrt{(11-1)^2 + (3 - (-2))^2 + (7 - (-8))^2} = \sqrt{100 + 25 + 225} = \sqrt{350} = 5\sqrt{14}\)
Since \(AB + BC = 2\sqrt{14} + 3\sqrt{14} = 5\sqrt{14} = AC\), the points are collinear.
Let \(B\) divide \(AC\) in the ratio \(k : 1\). Using the \(x\)-coordinate section formula:
\(\frac{11k + 1}{k+1} = 5 \implies 11k + 1 = 5k + 5 \implies 6k = 4 \implies k = \frac{4}{6} = \frac{2}{3}\).
Thus, \(B\) divides \(AC\) in the ratio \(2 : 3\) internally.
In simple words: Sum of two shorter segments equaling the longest segment proves collinearity.

Question 18. \(A\) is a point \((1, 3, 4)\) and \(B\) is the point \((1, -2, -1)\). A point \(P\) moves so that \(3PA = 2PB\). Find the locus of \(P\).
Answer:
Let the coordinates of \(P\) be \((x, y, z)\).
Given condition: \(3PA = 2PB \implies 9PA^2 = 4PB^2\).
Using distance squared formula:
\(9 \left[ (x-1)^2 + (y-3)^2 + (z-4)^2 \right] = 4 \left[ (x-1)^2 + (y - (-2))^2 + (z - (-1))^2 \right]\)
\(9 \left[ x^2 - 2x + 1 + y^2 - 6y + 9 + z^2 - 8z + 16 \right] = 4 \left[ x^2 - 2x + 1 + y^2 + 4y + 4 + z^2 + 2z + 1 \right]\)
\(9 \left[ x^2 + y^2 + z^2 - 2x - 6y - 8z + 26 \right] = 4 \left[ x^2 + y^2 + z^2 - 2x + 4y + 2z + 6 \right]\)
Expand and group terms:
\(9x^2 + 9y^2 + 9z^2 - 18x - 54y - 72z + 234 = 4x^2 + 4y^2 + 4z^2 - 8x + 16y + 8z + 24\)
Subtract RHS from LHS:
\((9-4)x^2 + (9-4)y^2 + (9-4)z^2 + (-18 - (-8))x + (-54 - 16)y + (-72 - 8)z + (234 - 24) = 0\)
\(5x^2 + 5y^2 + 5z^2 - 10x - 70y - 80z + 210 = 0\).
In simple words: Square both sides of the distance condition \(3PA = 2PB\) and simplify algebraic terms.

Exam Tip: This locus represents a sphere (Appollonius sphere in 3D).

Question 19. Show that the points \(A(0, 1, 2)\), \(B(2, -1, 3)\) and \(C(1, -3, 1)\) are vertices of an isosceles right-angled triangle.
Answer:
Calculate the squares of the side lengths:\
- \(AB^2 = (2-0)^2 + (-1-1)^2 + (3-2)^2 = 4 + 4 + 1 = 9 \implies AB = 3\)
- \(BC^2 = (1-2)^2 + (-3 - (-1))^2 + (1-3)^2 = 1 + 4 + 4 = 9 \implies BC = 3\)
- \(AC^2 = (1-0)^2 + (-3-1)^2 + (1-2)^2 = 1 + 16 + 1 = 18 \implies AC = \sqrt{18} = 3\sqrt{2}\)
Since \(AB = BC = 3\), the triangle is isosceles.
Check Pythagoras theorem: \(AB^2 + BC^2 = 9 + 9 = 18 = AC^2\).
Since \(AB^2 + BC^2 = AC^2\), it is a right-angled triangle at vertex \(B\).
In simple words: Two equal sides and satisfaction of Pythagoras theorem proves it is an isosceles right-angled triangle.

Exam Tip: Using squared distances avoids messy square root calculations.

Question 20. Show that the points \(A(1, 3, 4)\), \(B(-1, 6, 10)\), \(C(-7, 4, 7)\) and \(D(-5, 1, 1)\) are the vertices of a rhombus.
Answer:
Calculate all four side lengths:
- \(AB = \sqrt{(-1-1)^2 + (6-3)^2 + (10-4)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7\)
- \(BC = \sqrt{(-7 - (-1))^2 + (4-6)^2 + (7-10)^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7\)
- \(CD = \sqrt{(-5 - (-7))^2 + (1-4)^2 + (1-7)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7\)
- \(DA = \sqrt{(1 - (-5))^2 + (3-1)^2 + (4-1)^2} = \sqrt{36 + 4 + 9} = \sqrt{49} = 7\)
All four sides are equal: \(AB = BC = CD = DA = 7\).
Now check the diagonals:
- \(AC = \sqrt{(-7-1)^2 + (4-3)^2 + (7-4)^2} = \sqrt{64 + 1 + 9} = \sqrt{74}\)
- \(BD = \sqrt{(-5 - (-1))^2 + (1-6)^2 + (1-10)^2} = \sqrt{16 + 25 + 81} = \sqrt{122}\)
Since all four sides are equal but diagonals are unequal (\(AC \neq BD\)), the figure is a rhombus.
In simple words: A quadrilateral with four equal sides and unequal diagonals is a rhombus.

Exam Tip: A rhombus has all sides equal, unlike a parallelogram where only opposite sides are equal.

Question 21. Find the point on \(z\)-axis which is equidistant from \((1, 5, 7)\) and \((5, 1, -4)\).
Answer:
Any point on the \(z\)-axis has coordinates of the form \(P(0, 0, z)\).
Let this point be equidistant from \(A(1, 5, 7)\) and \(B(5, 1, -4)\). Therefore, \(PA^2 = PB^2\):
\((0-1)^2 + (0-5)^2 + (z-7)^2 = (0-5)^2 + (0-1)^2 + (z - (-4))^2\)
\(1 + 25 + (z^2 - 14z + 49) = 25 + 1 + (z^2 + 8z + 16)\)
Cancel common terms from both sides:
\(-14z + 75 = 8z + 42\)
\(75 - 42 = 8z + 14z\)
\(33 = 22z \implies z = \frac{33}{22} = \frac{3}{2}\).
Thus, the point is \((0, 0, \frac{3}{2})\).
In simple words: Points on the \(z\)-axis have \(x = 0\) and \(y = 0\). Equate squared distances.

Exam Tip: For points on axes, use coordinates like \((0,0,z)\) for \(z\)-axis, \((x,0,0)\) for \(x\)-axis, etc.

Question 22. Find the value of \(p\) such that distance between the points \((4, 5, p)\) and \((7, 1, -3)\) is \(13\).
Answer:
Using the distance formula:
\(\sqrt{(7-4)^2 + (1-5)^2 + (-3-p)^2} = 13\)
Square both sides:
\((3)^2 + (-4)^2 + (-3-p)^2 = 169\)
\(9 + 16 + (3+p)^2 = 169\)
\(25 + (3+p)^2 = 169\)
\((3+p)^2 = 169 - 25 = 144\)
Taking square root on both sides:
\(3 + p = \pm 12\)
- Case 1: \(3 + p = 12 \implies p = 9\)
- Case 2: \(3 + p = -12 \implies p = -15\)
Thus, \(p = 9\) or \(p = -15\).
In simple words: Set up the distance equation equal to \(13\), square both sides, and solve the quadratic in \(p\).

Question 23. Three vertices of a parallelogram ABCD are \(A(3, 4, 6)\), \(B(-1, 0, 6)\) and \(C(5, 2, 0)\). Find the coordinate of \(D\).
Answer:
Let the coordinates of \(D\) be \((x, y, z)\).
In a parallelogram, the diagonals bisect each other. Therefore, the midpoint of diagonal \(AC\) equals the midpoint of diagonal \(BD\):
Midpoint of \(AC = \left( \frac{3+5}{2}, \frac{4+2}{2}, \frac{6+0}{2} \right) = (4, 3, 3)\).
Midpoint of \(BD = \left( \frac{-1+x}{2}, \frac{0+y}{2}, \frac{6+z}{2} \right)\).
Equating corresponding coordinates:
\(\frac{-1+x}{2} = 4 \implies -1 + x = 8 \implies x = 9\)
\(\frac{0+y}{2} = 3 \implies y = 6\)
\(\frac{6+z}{2} = 3 \implies 6 + z = 6 \implies z = 0\)
Thus, the coordinates of \(D\) are \((9, 6, 0)\).
In simple words: Diagonals of a parallelogram share the exact same midpoint.

Exam Tip: Midpoint formula in 3D is \(\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2}\right)\).

Question 24. Find the distance of the point \((1, 2, 0)\) from the point where line joining \(A(2, -3, 1)\) and \(B(3, -4, -5)\) cuts the plane \(2x + y + z = 7\).
Answer:
The equation of the line passing through \(A(2, -3, 1)\) and \(B(3, -4, -5)\) is:
\(\frac{x - 2}{3 - 2} = \frac{y - (-3)}{-4 - (-3)} = \frac{z - 1}{-5 - 1} = k\)
\(\frac{x - 2}{1} = \frac{y + 3}{-1} = \frac{z - 1}{-6} = k\)
Any general point on this line is given by \((k+2, -k-3, -6k+1)\).
Since this point lies on the plane \(2x + y + z = 7\), substitute the coordinates into the plane equation:
\(2(k+2) + (-k-3) + (-6k+1) = 7\)
\(2k + 4 - k - 3 - 6k + 1 = 7\)
\(-5k + 2 = 7 \implies -5k = 5 \implies k = -1\).
Substitute \(k = -1\) to find the point of intersection \(P\):
\(P = (-1+2, -(-1)-3, -6(-1)+1) = (1, -2, 7)\).
Now, find the distance between \((1, 2, 0)\) and \(P(1, -2, 7)\):
Distance = \(\sqrt{(1-1)^2 + (-2 - 2)^2 + (7-0)^2} = \sqrt{0 + (-4)^2 + 7^2} = \sqrt{16 + 49} = \sqrt{65}\)... Wait, let's re-verify the calculation.
Let's check the plane equation and line intersection: \(2(1) + (-2) + 7 = 7\), correct.
Distance = \(\sqrt{0 + 16 + 49} = \sqrt{65}\). (Note: Answer key simplification check: source states \(\frac{\sqrt{281}}{3}\), let's re-check line coordinates if ratio was used. Using section formula, ratio \(k\) with plane gives \(-5k+2 = 7 \implies k = -1\). Distance = \(\sqrt{65}\)).
In simple words: Find parametric coordinates of the line, substitute into the plane equation to find the intersection point, then calculate distance.

Exam Tip: Parametric form of a line makes finding plane intersections straightforward.

Question 25. The midpoints of the sides of the triangle are \((1, 5, -1)\), \((0, 4, -2)\) and \((2, 3, 4)\). Find its vertices.
Answer:
Let the vertices of the triangle be \(A(x_1, y_1, z_1)\), \(B(x_2, y_2, z_2)\), and \(C(x_3, y_3, z_3)\).
Given midpoints are \(D(1, 5, -1)\), \(E(0, 4, -2)\), and \(F(2, 3, 4)\).
Using midpoint relations:
\(\frac{x_1 + x_2}{2} = 1 \implies x_1 + x_2 = 2\) -- (1)
\(\frac{x_2 + x_3}{2} = 0 \implies x_2 + x_3 = 0\) -- (2)
\(\frac{x_3 + x_1}{2} = 2 \implies x_3 + x_1 = 4\) -- (3)
Adding (1), (2), and (3): \(2(x_1 + x_2 + x_3) = 6 \implies x_1 + x_2 + x_3 = 3\).
- Subtracting (2): \(x_1 = 3 - 0 = 3\)
- Subtracting (3): \(x_2 = 3 - 4 = -1\)
- Subtracting (1): \(x_3 = 3 - 2 = 1\)
Similarly for \(y\)-coordinates:
\(y_1 + y_2 = 10\), \(y_2 + y_3 = 8\), \(y_3 + y_1 = 6 \implies y_1 + y_2 + y_3 = 12\).
\(y_1 = 12 - 8 = 4\), \(y_2 = 12 - 6 = 6\), \(y_3 = 12 - 10 = 2\).
Similarly for \(z\)-coordinates:
\(z_1 + z_2 = -2\), \(z_2 + z_3 = -4\), \(z_3 + z_1 = 8 \implies z_1 + z_2 + z_3 = 1\).
\(z_1 = 1 - (-4) = 5\), \(z_2 = 1 - 8 = -7\), \(z_3 = 1 - (-2) = 3\).
Thus, the vertices are \((3, 4, 5)\), \((-1, 6, -7)\), and \((1, 2, 3)\).
In simple words: Sum all midpoint equations to find the sum of all three coordinates, then subtract pairwise.

Exam Tip: This shortcut method works for any triangle given the midpoints of its sides.

Question 26. Determine the point in \(xy\)-plane which is equidistant from three points \(A(2, 0, 3)\), \(B(0, 3, 2)\) and \(C(0, 0, 1)\).
Answer:
Any point in the \(xy\)-plane has coordinates \((x, y, 0)\).
Let this point be \(P(x, y, 0)\). Given \(PA^2 = PB^2 = PC^2\).
- Equating \(PA^2 = PC^2\):
\((x-2)^2 + (y-0)^2 + (0-3)^2 = (x-0)^2 + (y-0)^2 + (0-1)^2\)
\(x^2 - 4x + 4 + y^2 + 9 = x^2 + y^2 + 1\)
\(-4x + 13 = 1 \implies 4x = 12 \implies x = 3\).
- Equating \(PB^2 = PC^2\):
\((x-0)^2 + (y-3)^2 + (0-2)^2 = x^2 + y^2 + 1\)
\(x^2 + y^2 - 6y + 9 + 4 = x^2 + y^2 + 1\)
\(-6y + 13 = 1 \implies 6y = 12 \implies y = 2\).
Thus, the point is \((3, 2, 0)\).
In simple words: Set \(z = 0\) for the \(xy\)-plane and equate squared distances to find \(x\) and \(y\).

Exam Tip: This point is the circumcenter of the triangle formed by the three points in space.