RBSE Solutions Class 12 Maths Chapter 13 Vector Exercise 13.5

Question 1. Find the value of \( \overrightarrow{a} \times (\overrightarrow{b} \times \overrightarrow{c}) \) if
(i) \( \overrightarrow{a} =3\hat{i}-\hat{j}+\hat{k} \), \( \overrightarrow{b} = \hat{i} +3\hat{j}+\hat{k} \) and \( \overrightarrow{c} =-\hat{i}+\hat{j}+3\hat{k} \)
(ii) \( \overrightarrow{a} = 2\hat{i}+\hat{j}-3\hat{k} \), \( \overrightarrow{b}=\hat{i}-2\hat{j}+\hat{k} \) and \( \overrightarrow{c} =-\hat{i}+\hat{j}-4\hat{k} \)
Answer:
(i) Given vectors are:
\( \overrightarrow{a} =3\hat{i}-\hat{j}+\hat{k} \)
\( \overrightarrow{b} = \hat{i} +3\hat{j}+\hat{k} \)
\( \overrightarrow{c} =-\hat{i}+\hat{j}+3\hat{k} \)
We use the vector triple product formula: \( \overrightarrow{a} \times (\overrightarrow{b} \times \overrightarrow{c}) = (\overrightarrow{a} \cdot \overrightarrow{c})\overrightarrow{b} - (\overrightarrow{a} \cdot \overrightarrow{b})\overrightarrow{c} \).
First, calculate the dot product \( \overrightarrow{a} \cdot \overrightarrow{c} \):
\( \overrightarrow{a} \cdot \overrightarrow{c} = (3\hat{i}-\hat{j}+\hat{k}) \cdot (-\hat{i}+\hat{j}+3\hat{k}) \)
\( = (3)(-1) + (-1)(1) + (1)(3) \)
\( = -3 - 1 + 3 \)
\( \implies \overrightarrow{a} \cdot \overrightarrow{c} = -1 \)
Next, calculate the dot product \( \overrightarrow{a} \cdot \overrightarrow{b} \):
\( \overrightarrow{a} \cdot \overrightarrow{b} = (3\hat{i}-\hat{j}+\hat{k}) \cdot (\hat{i}+3\hat{j}-\hat{k}) \)
\( = (3)(1) + (-1)(3) + (1)(-1) \)
\( = 3 - 3 - 1 \)
\( \implies \overrightarrow{a} \cdot \overrightarrow{b} = -1 \)
Now, substitute these values into the formula:
\( \overrightarrow{a} \times (\overrightarrow{b} \times \overrightarrow{c}) = (\overrightarrow{a} \cdot \overrightarrow{c})\overrightarrow{b} - (\overrightarrow{a} \cdot \overrightarrow{b})\overrightarrow{c} \)
\( = (-1)(\hat{i}+3\hat{j}+\hat{k}) - (-1)(-\hat{i}+\hat{j}+3\hat{k}) \)
\( = -\hat{i}-3\hat{j}-\hat{k} + (-\hat{i}+\hat{j}+3\hat{k}) \)
\( = -\hat{i}-3\hat{j}-\hat{k} - \hat{i}+\hat{j}+3\hat{k} \)
\( \implies \overrightarrow{a} \times (\overrightarrow{b} \times \overrightarrow{c}) = -2\hat{i}-2\hat{j}+2\hat{k} \)

(ii) Given vectors are:
\( \overrightarrow{a} = 2\hat{i}+\hat{j}-3\hat{k} \)
\( \overrightarrow{b}=\hat{i}-2\hat{j}+\hat{k} \)
\( \overrightarrow{c} =-\hat{i}+\hat{j}-4\hat{k} \)
Again, use the vector triple product formula: \( \overrightarrow{a} \times (\overrightarrow{b} \times \overrightarrow{c}) = (\overrightarrow{a} \cdot \overrightarrow{c})\overrightarrow{b} - (\overrightarrow{a} \cdot \overrightarrow{b})\overrightarrow{c} \).
First, calculate the dot product \( \overrightarrow{a} \cdot \overrightarrow{c} \):
\( \overrightarrow{a} \cdot \overrightarrow{c} = (2\hat{i}+\hat{j}-3\hat{k}) \cdot (-\hat{i}+\hat{j}-4\hat{k}) \)
\( = (2)(-1) + (1)(1) + (-3)(-4) \)
\( = -2 + 1 + 12 \)
\( \implies \overrightarrow{a} \cdot \overrightarrow{c} = 11 \)
Next, calculate the dot product \( \overrightarrow{a} \cdot \overrightarrow{b} \):
\( \overrightarrow{a} \cdot \overrightarrow{b} = (2\hat{i}+\hat{j}-3\hat{k}) \cdot (\hat{i}-2\hat{j}+\hat{k}) \)
\( = (2)(1) + (1)(-2) + (-3)(1) \)
\( = 2 - 2 - 3 \)
\( \implies \overrightarrow{a} \cdot \overrightarrow{b} = -3 \)
Now, substitute these values into the formula:
\( \overrightarrow{a} \times (\overrightarrow{b} \times \overrightarrow{c}) = (\overrightarrow{a} \cdot \overrightarrow{c})\overrightarrow{b} - (\overrightarrow{a} \cdot \overrightarrow{b})\overrightarrow{c} \)
\( = (11)(\hat{i}-2\hat{j}+\hat{k}) - (-3)(-\hat{i}+\hat{j}-4\hat{k}) \)
\( = 11\hat{i}-22\hat{j}+11\hat{k} + 3(-\hat{i}+\hat{j}-4\hat{k}) \)
\( = 11\hat{i}-22\hat{j}+11\hat{k} - 3\hat{i}+3\hat{j}-12\hat{k} \)
\( \implies \overrightarrow{a} \times (\overrightarrow{b} \times \overrightarrow{c}) = 8\hat{i}-19\hat{j}-\hat{k} \)
In simple words: To find this vector, we use a special formula that involves calculating dot products first. Then we substitute those values back into the formula and simplify to get the final vector. This makes complex vector calculations easier.
🎯 Exam Tip: Remember the vector triple product formula: \( \overrightarrow{a} \times (\overrightarrow{b} \times \overrightarrow{c}) = (\overrightarrow{a} \cdot \overrightarrow{c})\overrightarrow{b} - (\overrightarrow{a} \cdot \overrightarrow{b})\overrightarrow{c} \). Be careful with signs during dot product and final subtraction.

Question 2. Prove that \( \overrightarrow{a} \times (\overrightarrow{b} \times \overrightarrow{c}) \neq (\overrightarrow{a} \times \overrightarrow{b}) \times \overrightarrow{c} \) if
(i) \( \overrightarrow{a} = 2\hat{i}+5\hat{j}-7\hat{k} \), \( \overrightarrow{b} = -3\hat{i}+4\hat{j}+\hat{k} \), and \( \overrightarrow{c} =-\hat{i}-2\hat{j}-3\hat{k} \)
(ii) \( \overrightarrow{a} = 2\hat{i}+3\hat{j}-5\hat{k} \), \( \overrightarrow{b} = -\hat{i}+\hat{j}+\sqrt{2}\hat{k} \), and \( \overrightarrow{c} =4\hat{i}-2\hat{j}+\sqrt{3}\hat{k} \)
Answer:
(i) Given vectors are:
\( \overrightarrow{a} = 2\hat{i}+5\hat{j}-7\hat{k} \)
\( \overrightarrow{b} = -3\hat{i}+4\hat{j}+\hat{k} \)
\( \overrightarrow{c} =-\hat{i}-2\hat{j}-3\hat{k} \)
We need to prove \( \overrightarrow{a} \times (\overrightarrow{b} \times \overrightarrow{c}) \neq (\overrightarrow{a} \times \overrightarrow{b}) \times \overrightarrow{c} \).

**First, calculate \( \overrightarrow{a} \times (\overrightarrow{b} \times \overrightarrow{c}) \):**
Using the formula \( \overrightarrow{a} \times (\overrightarrow{b} \times \overrightarrow{c}) = (\overrightarrow{a} \cdot \overrightarrow{c})\overrightarrow{b} - (\overrightarrow{a} \cdot \overrightarrow{b})\overrightarrow{c} \).
Calculate \( \overrightarrow{a} \cdot \overrightarrow{c} \):
\( \overrightarrow{a} \cdot \overrightarrow{c} = (2\hat{i}+5\hat{j}-7\hat{k}) \cdot (-\hat{i}-2\hat{j}-3\hat{k}) \)
\( = (2)(-1) + (5)(-2) + (-7)(-3) \)
\( = -2 - 10 + 21 \)
\( \implies \overrightarrow{a} \cdot \overrightarrow{c} = 9 \)
Calculate \( \overrightarrow{a} \cdot \overrightarrow{b} \):
\( \overrightarrow{a} \cdot \overrightarrow{b} = (2\hat{i}+5\hat{j}-7\hat{k}) \cdot (-3\hat{i}+4\hat{j}+\hat{k}) \)
\( = (2)(-3) + (5)(4) + (-7)(1) \)
\( = -6 + 20 - 7 \)
\( \implies \overrightarrow{a} \cdot \overrightarrow{b} = 7 \)
Substitute into the formula:
\( \overrightarrow{a} \times (\overrightarrow{b} \times \overrightarrow{c}) = (9)\overrightarrow{b} - (7)\overrightarrow{c} \)
\( = 9(-3\hat{i}+4\hat{j}+\hat{k}) - 7(-\hat{i}-2\hat{j}-3\hat{k}) \)
\( = -27\hat{i}+36\hat{j}+9\hat{k} + 7\hat{i}+14\hat{j}+21\hat{k} \)
\( \implies \overrightarrow{a} \times (\overrightarrow{b} \times \overrightarrow{c}) = -20\hat{i}+50\hat{j}+30\hat{k} \) (Equation 1)

**Next, calculate \( (\overrightarrow{a} \times \overrightarrow{b}) \times \overrightarrow{c} \):**
Using the formula \( (\overrightarrow{a} \times \overrightarrow{b}) \times \overrightarrow{c} = (\overrightarrow{a} \cdot \overrightarrow{c})\overrightarrow{b} - (\overrightarrow{b} \cdot \overrightarrow{c})\overrightarrow{a} \).
We already have \( \overrightarrow{a} \cdot \overrightarrow{c} = 9 \).
Calculate \( \overrightarrow{b} \cdot \overrightarrow{c} \):
\( \overrightarrow{b} \cdot \overrightarrow{c} = (-3\hat{i}+4\hat{j}+\hat{k}) \cdot (-\hat{i}-2\hat{j}-3\hat{k}) \)
\( = (-3)(-1) + (4)(-2) + (1)(-3) \)
\( = 3 - 8 - 3 \)
\( \implies \overrightarrow{b} \cdot \overrightarrow{c} = -8 \)
Substitute into the formula:
\( (\overrightarrow{a} \times \overrightarrow{b}) \times \overrightarrow{c} = (9)\overrightarrow{b} - (-8)\overrightarrow{a} \)
\( = 9(-3\hat{i}+4\hat{j}+\hat{k}) + 8(2\hat{i}+5\hat{j}-7\hat{k}) \)
\( = -27\hat{i}+36\hat{j}+9\hat{k} + 16\hat{i}+40\hat{j}-56\hat{k} \)
\( \implies (\overrightarrow{a} \times \overrightarrow{b}) \times \overrightarrow{c} = -11\hat{i}+76\hat{j}-47\hat{k} \) (Equation 2)
Comparing Equation (1) and (2), we see that \( -20\hat{i}+50\hat{j}+30\hat{k} \neq -11\hat{i}+76\hat{j}-47\hat{k} \).
Hence, \( \overrightarrow{a} \times (\overrightarrow{b} \times \overrightarrow{c}) \neq (\overrightarrow{a} \times \overrightarrow{b}) \times \overrightarrow{c} \). Proven.

(ii) Given vectors are:
\( \overrightarrow{a} = 2\hat{i}+3\hat{j}-5\hat{k} \)
\( \overrightarrow{b} = -\hat{i}+\hat{j}+\sqrt{2}\hat{k} \)
\( \overrightarrow{c} =4\hat{i}-2\hat{j}+\sqrt{3}\hat{k} \)
We need to prove \( \overrightarrow{a} \times (\overrightarrow{b} \times \overrightarrow{c}) \neq (\overrightarrow{a} \times \overrightarrow{b}) \times \overrightarrow{c} \).

**First, calculate \( \overrightarrow{a} \times (\overrightarrow{b} \times \overrightarrow{c}) \):**
Using the formula \( \overrightarrow{a} \times (\overrightarrow{b} \times \overrightarrow{c}) = (\overrightarrow{a} \cdot \overrightarrow{c})\overrightarrow{b} - (\overrightarrow{a} \cdot \overrightarrow{b})\overrightarrow{c} \).
Calculate \( \overrightarrow{a} \cdot \overrightarrow{c} \):
\( \overrightarrow{a} \cdot \overrightarrow{c} = (2\hat{i}+3\hat{j}-5\hat{k}) \cdot (4\hat{i}-2\hat{j}+\sqrt{3}\hat{k}) \)
\( = (2)(4) + (3)(-2) + (-5)(\sqrt{3}) \)
\( = 8 - 6 - 5\sqrt{3} \)
\( \implies \overrightarrow{a} \cdot \overrightarrow{c} = 2 - 5\sqrt{3} \)
Calculate \( \overrightarrow{a} \cdot \overrightarrow{b} \):
\( \overrightarrow{a} \cdot \overrightarrow{b} = (2\hat{i}+3\hat{j}-5\hat{k}) \cdot (-\hat{i}+\hat{j}+\sqrt{2}\hat{k}) \)
\( = (2)(-1) + (3)(1) + (-5)(\sqrt{2}) \)
\( = -2 + 3 - 5\sqrt{2} \)
\( \implies \overrightarrow{a} \cdot \overrightarrow{b} = 1 - 5\sqrt{2} \)
Substitute into the formula:
\( \overrightarrow{a} \times (\overrightarrow{b} \times \overrightarrow{c}) = (2-5\sqrt{3})\overrightarrow{b} - (1-5\sqrt{2})\overrightarrow{c} \)
\( = (2-5\sqrt{3})(-\hat{i}+\hat{j}+\sqrt{2}\hat{k}) - (1-5\sqrt{2})(4\hat{i}-2\hat{j}+\sqrt{3}\hat{k}) \)
\( = (-2+5\sqrt{3})\hat{i} + (2-5\sqrt{3})\hat{j} + (2\sqrt{2}-5\sqrt{6})\hat{k} \)
\( - (4-20\sqrt{2})\hat{i} - (-2+10\sqrt{2})\hat{j} - (\sqrt{3}-5\sqrt{6})\hat{k} \)
\( = (-2+5\sqrt{3}-4+20\sqrt{2})\hat{i} + (2-5\sqrt{3}+2-10\sqrt{2})\hat{j} + (2\sqrt{2}-5\sqrt{6}-\sqrt{3}+5\sqrt{6})\hat{k} \)
\( \implies \overrightarrow{a} \times (\overrightarrow{b} \times \overrightarrow{c}) = (-6+5\sqrt{3}+20\sqrt{2})\hat{i} + (4-5\sqrt{3}-10\sqrt{2})\hat{j} + (2\sqrt{2}-\sqrt{3})\hat{k} \) (Equation 1)

**Next, calculate \( (\overrightarrow{a} \times \overrightarrow{b}) \times \overrightarrow{c} \):**
Using the formula \( (\overrightarrow{a} \times \overrightarrow{b}) \times \overrightarrow{c} = (\overrightarrow{a} \cdot \overrightarrow{c})\overrightarrow{b} - (\overrightarrow{b} \cdot \overrightarrow{c})\overrightarrow{a} \).
We already have \( \overrightarrow{a} \cdot \overrightarrow{c} = 2-5\sqrt{3} \).
Calculate \( \overrightarrow{b} \cdot \overrightarrow{c} \):
\( \overrightarrow{b} \cdot \overrightarrow{c} = (-\hat{i}+\hat{j}+\sqrt{2}\hat{k}) \cdot (4\hat{i}-2\hat{j}+\sqrt{3}\hat{k}) \)
\( = (-1)(4) + (1)(-2) + (\sqrt{2})(\sqrt{3}) \)
\( = -4 - 2 + \sqrt{6} \)
\( \implies \overrightarrow{b} \cdot \overrightarrow{c} = -6+\sqrt{6} \)
Substitute into the formula:
\( (\overrightarrow{a} \times \overrightarrow{b}) \times \overrightarrow{c} = (2-5\sqrt{3})\overrightarrow{b} - (-6+\sqrt{6})\overrightarrow{a} \)
\( = (2-5\sqrt{3})(-\hat{i}+\hat{j}+\sqrt{2}\hat{k}) - (-6+\sqrt{6})(2\hat{i}+3\hat{j}-5\hat{k}) \)
\( = (-2+5\sqrt{3})\hat{i} + (2-5\sqrt{3})\hat{j} + (2\sqrt{2}-5\sqrt{6})\hat{k} \)
\( - (-12+2\sqrt{6})\hat{i} - (-18+3\sqrt{6})\hat{j} - (30-5\sqrt{6})\hat{k} \)
\( = (-2+5\sqrt{3}+12-2\sqrt{6})\hat{i} + (2-5\sqrt{3}+18-3\sqrt{6})\hat{j} + (2\sqrt{2}-5\sqrt{6}-30+5\sqrt{6})\hat{k} \)
\( \implies (\overrightarrow{a} \times \overrightarrow{b}) \times \overrightarrow{c} = (10+5\sqrt{3}-2\sqrt{6})\hat{i} + (20-5\sqrt{3}-3\sqrt{6})\hat{j} + (2\sqrt{2}-30)\hat{k} \) (Equation 2)
Comparing Equation (1) and (2), we see that they are not equal.
Hence, \( \overrightarrow{a} \times (\overrightarrow{b} \times \overrightarrow{c}) \neq (\overrightarrow{a} \times \overrightarrow{b}) \times \overrightarrow{c} \). Proven.
In simple words: To prove that the order of cross products matters, we calculate both sides of the inequality separately. We use the vector triple product expansion formula and carefully compute all dot products and vector subtractions. Since the final vector results are different, we show the inequality holds true.
🎯 Exam Tip: Remember that the vector cross product is not associative, meaning \( \overrightarrow{a} \times (\overrightarrow{b} \times \overrightarrow{c}) \neq (\overrightarrow{a} \times \overrightarrow{b}) \times \overrightarrow{c} \). Be meticulous with calculations and signs, especially when dealing with square roots.

Question 3. Evaluate the formula \( \overrightarrow{a} \times (\overrightarrow{b} \times \overrightarrow{c}) = (\overrightarrow{a} \cdot \overrightarrow{c}) \overrightarrow{b} - (\overrightarrow{a} \cdot \overrightarrow{b}) \overrightarrow{c} \) if
(i) \( \overrightarrow{a} = \hat{i}+\hat{j}-2\hat{k} \), \( \overrightarrow{b}=2\hat{i}-\hat{j}+\hat{k} \), \( \overrightarrow{c} = \hat{i} +3\hat{j}-\hat{k} \)
(ii) \( \overrightarrow{a} = \hat{i}-2\hat{j}+\hat{k} \), \( \overrightarrow{b}=2\hat{i}+\hat{j}-\hat{k} \), \( \overrightarrow{c}=3\hat{i}+5\hat{j}+2\hat{k} \)
Answer:
(i) Given vectors are:
\( \overrightarrow{a} = \hat{i}+\hat{j}-2\hat{k} \)
\( \overrightarrow{b}=2\hat{i}-\hat{j}+\hat{k} \)
\( \overrightarrow{c} = \hat{i} +3\hat{j}-\hat{k} \)
We need to evaluate both sides of the formula \( \overrightarrow{a} \times (\overrightarrow{b} \times \overrightarrow{c}) = (\overrightarrow{a} \cdot \overrightarrow{c}) \overrightarrow{b} - (\overrightarrow{a} \cdot \overrightarrow{b}) \overrightarrow{c} \).

**Calculate the left-hand side (L.H.S.): \( \overrightarrow{a} \times (\overrightarrow{b} \times \overrightarrow{c}) \)**
First, find \( \overrightarrow{b} \times \overrightarrow{c} \):
\[ \overrightarrow{b} \times \overrightarrow{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 1 & 3 & -1 \end{vmatrix} \] \( = \hat{i}((-1)(-1) - (1)(3)) - \hat{j}((2)(-1) - (1)(1)) + \hat{k}((2)(3) - (-1)(1)) \)
\( = \hat{i}(1 - 3) - \hat{j}(-2 - 1) + \hat{k}(6 + 1) \)
\( = -2\hat{i} + 3\hat{j} + 7\hat{k} \)
Next, find \( \overrightarrow{a} \times (\overrightarrow{b} \times \overrightarrow{c}) \):
\[ \overrightarrow{a} \times (\overrightarrow{b} \times \overrightarrow{c}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -2 \\ -2 & 3 & 7 \end{vmatrix} \] \( = \hat{i}((1)(7) - (-2)(3)) - \hat{j}((1)(7) - (-2)(-2)) + \hat{k}((1)(3) - (1)(-2)) \)
\( = \hat{i}(7 + 6) - \hat{j}(7 - 4) + \hat{k}(3 + 2) \)
\( = 13\hat{i} - 3\hat{j} + 5\hat{k} \) (Equation 1)

**Calculate the right-hand side (R.H.S.): \( (\overrightarrow{a} \cdot \overrightarrow{c}) \overrightarrow{b} - (\overrightarrow{a} \cdot \overrightarrow{b}) \overrightarrow{c} \)**
First, find \( \overrightarrow{a} \cdot \overrightarrow{c} \):
\( \overrightarrow{a} \cdot \overrightarrow{c} = (\hat{i}+\hat{j}-2\hat{k}) \cdot (\hat{i}+3\hat{j}-\hat{k}) \)
\( = (1)(1) + (1)(3) + (-2)(-1) \)
\( = 1 + 3 + 2 \)
\( \implies \overrightarrow{a} \cdot \overrightarrow{c} = 6 \)
Next, find \( \overrightarrow{a} \cdot \overrightarrow{b} \):
\( \overrightarrow{a} \cdot \overrightarrow{b} = (\hat{i}+\hat{j}-2\hat{k}) \cdot (2\hat{i}-\hat{j}+\hat{k}) \)
\( = (1)(2) + (1)(-1) + (-2)(1) \)
\( = 2 - 1 - 2 \)
\( \implies \overrightarrow{a} \cdot \overrightarrow{b} = -1 \)
Substitute into the R.H.S. expression:
\( (\overrightarrow{a} \cdot \overrightarrow{c}) \overrightarrow{b} - (\overrightarrow{a} \cdot \overrightarrow{b}) \overrightarrow{c} = (6)(2\hat{i}-\hat{j}+\hat{k}) - (-1)(\hat{i}+3\hat{j}-\hat{k}) \)
\( = 12\hat{i}-6\hat{j}+6\hat{k} + \hat{i}+3\hat{j}-\hat{k} \)
\( = 13\hat{i}-3\hat{j}+5\hat{k} \) (Equation 2)
Comparing Equation (1) and Equation (2), we see that L.H.S. = R.H.S.
Thus, the formula is evaluated and shown to be true for these vectors.

(ii) Given vectors are:
\( \overrightarrow{a} = \hat{i}-2\hat{j}+\hat{k} \)
\( \overrightarrow{b}=2\hat{i}+\hat{j}-\hat{k} \)
\( \overrightarrow{c}=3\hat{i}+5\hat{j}+2\hat{k} \)
We need to evaluate both sides of the formula \( \overrightarrow{a} \times (\overrightarrow{b} \times \overrightarrow{c}) = (\overrightarrow{a} \cdot \overrightarrow{c}) \overrightarrow{b} - (\overrightarrow{a} \cdot \overrightarrow{b}) \overrightarrow{c} \).

**Calculate the left-hand side (L.H.S.): \( \overrightarrow{a} \times (\overrightarrow{b} \times \overrightarrow{c}) \)**
First, find \( \overrightarrow{b} \times \overrightarrow{c} \):
\[ \overrightarrow{b} \times \overrightarrow{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 3 & 5 & 2 \end{vmatrix} \] \( = \hat{i}((1)(2) - (-1)(5)) - \hat{j}((2)(2) - (-1)(3)) + \hat{k}((2)(5) - (1)(3)) \)
\( = \hat{i}(2 + 5) - \hat{j}(4 + 3) + \hat{k}(10 - 3) \)
\( = 7\hat{i} - 7\hat{j} + 7\hat{k} \)
Next, find \( \overrightarrow{a} \times (\overrightarrow{b} \times \overrightarrow{c}) \):
\[ \overrightarrow{a} \times (\overrightarrow{b} \times \overrightarrow{c}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -2 & 1 \\ 7 & -7 & 7 \end{vmatrix} \] \( = \hat{i}((-2)(7) - (1)(-7)) - \hat{j}((1)(7) - (1)(7)) + \hat{k}((1)(-7) - (-2)(7)) \)
\( = \hat{i}(-14 + 7) - \hat{j}(7 - 7) + \hat{k}(-7 + 14) \)
\( = -7\hat{i} + 0\hat{j} + 7\hat{k} \)
\( = -7\hat{i} + 7\hat{k} \) (Equation 1)

**Calculate the right-hand side (R.H.S.): \( (\overrightarrow{a} \cdot \overrightarrow{c}) \overrightarrow{b} - (\overrightarrow{a} \cdot \overrightarrow{b}) \overrightarrow{c} \)**
First, find \( \overrightarrow{a} \cdot \overrightarrow{c} \):
\( \overrightarrow{a} \cdot \overrightarrow{c} = (\hat{i}-2\hat{j}+\hat{k}) \cdot (3\hat{i}+5\hat{j}+2\hat{k}) \)
\( = (1)(3) + (-2)(5) + (1)(2) \)
\( = 3 - 10 + 2 \)
\( \implies \overrightarrow{a} \cdot \overrightarrow{c} = -5 \)
Next, find \( \overrightarrow{a} \cdot \overrightarrow{b} \):
\( \overrightarrow{a} \cdot \overrightarrow{b} = (\hat{i}-2\hat{j}+\hat{k}) \cdot (2\hat{i}+\hat{j}-\hat{k}) \)
\( = (1)(2) + (-2)(1) + (1)(-1) \)
\( = 2 - 2 - 1 \)
\( \implies \overrightarrow{a} \cdot \overrightarrow{b} = -1 \)
Substitute into the R.H.S. expression:
\( (\overrightarrow{a} \cdot \overrightarrow{c}) \overrightarrow{b} - (\overrightarrow{a} \cdot \overrightarrow{b}) \overrightarrow{c} = (-5)(2\hat{i}+\hat{j}-\hat{k}) - (-1)(3\hat{i}+5\hat{j}+2\hat{k}) \)
\( = -10\hat{i}-5\hat{j}+5\hat{k} + 3\hat{i}+5\hat{j}+2\hat{k} \)
\( = (-10+3)\hat{i} + (-5+5)\hat{j} + (5+2)\hat{k} \)
\( = -7\hat{i} + 0\hat{j} + 7\hat{k} \)
\( = -7\hat{i} + 7\hat{k} \) (Equation 2)
Comparing Equation (1) and Equation (2), we see that L.H.S. = R.H.S.
Thus, the formula is evaluated and shown to be true for these vectors.
In simple words: This question asks us to check if a known vector identity holds true for specific given vectors. We calculate both sides of the equation separately. First, we find the cross product of two vectors, then cross it with the third. For the other side, we find dot products and use them to scale and combine the original vectors. If both calculations lead to the exact same result, the formula is confirmed for those vectors.
🎯 Exam Tip: When evaluating vector identities, always compute both the L.H.S. and R.H.S. separately and ensure they match. This confirms the identity and helps catch calculation errors.

Question 4. For any vector \( \overrightarrow{a} \), prove that: \( \hat{i} \times (\overrightarrow{a} \times \hat{i}) + \hat{j} \times (\overrightarrow{a} \times \hat{j}) + \hat{k} \times (\overrightarrow{a} \times \hat{k}) = 2\overrightarrow{a} \)
Answer:
We use the vector triple product identity: \( \overrightarrow{A} \times (\overrightarrow{B} \times \overrightarrow{C}) = (\overrightarrow{A} \cdot \overrightarrow{C})\overrightarrow{B} - (\overrightarrow{A} \cdot \overrightarrow{B})\overrightarrow{C} \).
Let's apply this identity to each term in the given equation:
1. For the first term \( \hat{i} \times (\overrightarrow{a} \times \hat{i}) \):
Here, \( \overrightarrow{A} = \hat{i} \), \( \overrightarrow{B} = \overrightarrow{a} \), \( \overrightarrow{C} = \hat{i} \).
So, \( \hat{i} \times (\overrightarrow{a} \times \hat{i}) = (\hat{i} \cdot \hat{i})\overrightarrow{a} - (\hat{i} \cdot \overrightarrow{a})\hat{i} \).
Since \( \hat{i} \cdot \hat{i} = |\hat{i}|^2 = 1^2 = 1 \),
\( \hat{i} \times (\overrightarrow{a} \times \hat{i}) = (1)\overrightarrow{a} - (\hat{i} \cdot \overrightarrow{a})\hat{i} = \overrightarrow{a} - (\hat{i} \cdot \overrightarrow{a})\hat{i} \).
2. For the second term \( \hat{j} \times (\overrightarrow{a} \times \hat{j}) \):
Here, \( \overrightarrow{A} = \hat{j} \), \( \overrightarrow{B} = \overrightarrow{a} \), \( \overrightarrow{C} = \hat{j} \).
So, \( \hat{j} \times (\overrightarrow{a} \times \hat{j}) = (\hat{j} \cdot \hat{j})\overrightarrow{a} - (\hat{j} \cdot \overrightarrow{a})\hat{j} \).
Since \( \hat{j} \cdot \hat{j} = |\hat{j}|^2 = 1^2 = 1 \),
\( \hat{j} \times (\overrightarrow{a} \times \hat{j}) = (1)\overrightarrow{a} - (\hat{j} \cdot \overrightarrow{a})\hat{j} = \overrightarrow{a} - (\hat{j} \cdot \overrightarrow{a})\hat{j} \).
3. For the third term \( \hat{k} \times (\overrightarrow{a} \times \hat{k}) \):
Here, \( \overrightarrow{A} = \hat{k} \), \( \overrightarrow{B} = \overrightarrow{a} \), \( \overrightarrow{C} = \hat{k} \).
So, \( \hat{k} \times (\overrightarrow{a} \times \hat{k}) = (\hat{k} \cdot \hat{k})\overrightarrow{a} - (\hat{k} \cdot \overrightarrow{a})\hat{k} \).
Since \( \hat{k} \cdot \hat{k} = |\hat{k}|^2 = 1^2 = 1 \),
\( \hat{k} \times (\overrightarrow{a} \times \hat{k}) = (1)\overrightarrow{a} - (\hat{k} \cdot \overrightarrow{a})\hat{k} = \overrightarrow{a} - (\hat{k} \cdot \overrightarrow{a})\hat{k} \).
Now, we add all three expanded terms:
\( \hat{i} \times (\overrightarrow{a} \times \hat{i}) + \hat{j} \times (\overrightarrow{a} \times \hat{j}) + \hat{k} \times (\overrightarrow{a} \times \hat{k}) \)
\( = (\overrightarrow{a} - (\hat{i} \cdot \overrightarrow{a})\hat{i}) + (\overrightarrow{a} - (\hat{j} \cdot \overrightarrow{a})\hat{j}) + (\overrightarrow{a} - (\hat{k} \cdot \overrightarrow{a})\hat{k}) \)
\( = 3\overrightarrow{a} - [(\hat{i} \cdot \overrightarrow{a})\hat{i} + (\hat{j} \cdot \overrightarrow{a})\hat{j} + (\hat{k} \cdot \overrightarrow{a})\hat{k}] \).
Let \( \overrightarrow{a} \) be represented in component form as \( \overrightarrow{a} = l\hat{i} + m\hat{j} + n\hat{k} \).
Then:
\( \hat{i} \cdot \overrightarrow{a} = \hat{i} \cdot (l\hat{i} + m\hat{j} + n\hat{k}) = l(\hat{i} \cdot \hat{i}) + m(\hat{i} \cdot \hat{j}) + n(\hat{i} \cdot \hat{k}) = l(1) + m(0) + n(0) = l \).
Similarly, \( \hat{j} \cdot \overrightarrow{a} = m \) and \( \hat{k} \cdot \overrightarrow{a} = n \).
Substituting these values back into the expression:
\( (\hat{i} \cdot \overrightarrow{a})\hat{i} + (\hat{j} \cdot \overrightarrow{a})\hat{j} + (\hat{k} \cdot \overrightarrow{a})\hat{k} = l\hat{i} + m\hat{j} + n\hat{k} \).
This is exactly equal to \( \overrightarrow{a} \).
So, the expression becomes:
\( 3\overrightarrow{a} - \overrightarrow{a} = 2\overrightarrow{a} \).
Hence, \( \hat{i} \times (\overrightarrow{a} \times \hat{i}) + \hat{j} \times (\overrightarrow{a} \times \hat{j}) + \hat{k} \times (\overrightarrow{a} \times \hat{k}) = 2\overrightarrow{a} \). Proven.
In simple words: We use a special rule for vector triple products to break down each part of the equation. We notice that the dot product of a unit vector with itself is 1, and the dot product of any vector with a unit vector gives its component in that direction. When we add everything up, all the component terms cancel out except for twice the original vector itself. This shows a neat relationship between a vector and its cross products with the standard unit vectors.
🎯 Exam Tip: Master the vector triple product identity: \( \overrightarrow{A} \times (\overrightarrow{B} \times \overrightarrow{C}) = (\overrightarrow{A} \cdot \overrightarrow{C})\overrightarrow{B} - (\overrightarrow{A} \cdot \overrightarrow{B})\overrightarrow{C} \). Remember that \( \hat{i} \cdot \hat{i} = 1 \) and \( \hat{i} \cdot \hat{j} = 0 \), and \( \overrightarrow{a} = (\overrightarrow{a} \cdot \hat{i})\hat{i} + (\overrightarrow{a} \cdot \hat{j})\hat{j} + (\overrightarrow{a} \cdot \hat{k})\hat{k} \).

Question 5. Prove that
Answer:
This question is incomplete as the statement to be proven is missing. Please provide the full question.
In simple words: We cannot answer this question because the statement that needs to be proven is missing from the question text.
🎯 Exam Tip: Always read the full question carefully to ensure all necessary information is present before attempting to solve it. If a question is incomplete, politely state that it cannot be answered.

Question 6. Prove that \( \overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c} \) are coplanar if \( \overrightarrow{a} \times \overrightarrow{b}, \overrightarrow{b} \times \overrightarrow{c}, \overrightarrow{c} \times \overrightarrow{a} \) are coplanar.
Answer:
We are given that the vectors \( \overrightarrow{a} \times \overrightarrow{b} \), \( \overrightarrow{b} \times \overrightarrow{c} \), and \( \overrightarrow{c} \times \overrightarrow{a} \) are coplanar.
If three vectors are coplanar, their scalar triple product is zero.
So, \( [\overrightarrow{a} \times \overrightarrow{b}, \overrightarrow{b} \times \overrightarrow{c}, \overrightarrow{c} \times \overrightarrow{a}] = 0 \).
We know the identity for the scalar triple product of three cross products:
\( [\overrightarrow{A} \times \overrightarrow{B}, \overrightarrow{B} \times \overrightarrow{C}, \overrightarrow{C} \times \overrightarrow{A}] = [\overrightarrow{A}, \overrightarrow{B}, \overrightarrow{C}]^2 \).
Using this identity with \( \overrightarrow{A}=\overrightarrow{a} \), \( \overrightarrow{B}=\overrightarrow{b} \), \( \overrightarrow{C}=\overrightarrow{c} \), we get:
\( [\overrightarrow{a} \times \overrightarrow{b}, \overrightarrow{b} \times \overrightarrow{c}, \overrightarrow{c} \times \overrightarrow{a}] = [\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}]^2 \).
Since we know that \( [\overrightarrow{a} \times \overrightarrow{b}, \overrightarrow{b} \times \overrightarrow{c}, \overrightarrow{c} \times \overrightarrow{a}] = 0 \),
\( \implies [\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}]^2 = 0 \).
Taking the square root of both sides,
\( \implies [\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}] = 0 \).
The scalar triple product \( [\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}] \) being zero means that the vectors \( \overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c} \) are coplanar.
Thus, if \( \overrightarrow{a} \times \overrightarrow{b}, \overrightarrow{b} \times \overrightarrow{c}, \overrightarrow{c} \times \overrightarrow{a} \) are coplanar, then \( \overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c} \) are also coplanar. This proves the statement.
In simple words: This proof uses a special rule that relates the coplanarity of original vectors to the coplanarity of their cross products. If the cross products of three vectors are in the same plane (coplanar), then their scalar triple product is zero. A known identity shows that this directly means the scalar triple product of the original three vectors is also zero, which proves they are coplanar too.
🎯 Exam Tip: Remember that vectors \( \overrightarrow{A}, \overrightarrow{B}, \overrightarrow{C} \) are coplanar if and only if their scalar triple product \( [\overrightarrow{A}, \overrightarrow{B}, \overrightarrow{C}] = 0 \). Also, know the identity \( [\overrightarrow{a} \times \overrightarrow{b}, \overrightarrow{b} \times \overrightarrow{c}, \overrightarrow{c} \times \overrightarrow{a}] = [\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}]^2 \).

Question 7. Prove that \( (\overrightarrow{a} \times \overrightarrow{b}) \times (\overrightarrow{c} \times \overrightarrow{d}) = [\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]\overrightarrow{d} - [\overrightarrow{a} \overrightarrow{b} \overrightarrow{d}]\overrightarrow{c} \)
Answer:
Let L.H.S. \( = (\overrightarrow{a} \times \overrightarrow{b}) \times (\overrightarrow{c} \times \overrightarrow{d}) \).
Let's consider \( \overrightarrow{r} = \overrightarrow{a} \times \overrightarrow{b} \).
Then L.H.S. becomes \( \overrightarrow{r} \times (\overrightarrow{c} \times \overrightarrow{d}) \).
Using the vector triple product identity \( \overrightarrow{X} \times (\overrightarrow{Y} \times \overrightarrow{Z}) = (\overrightarrow{X} \cdot \overrightarrow{Z})\overrightarrow{Y} - (\overrightarrow{X} \cdot \overrightarrow{Y})\overrightarrow{Z} \),
we can write:
\( \overrightarrow{r} \times (\overrightarrow{c} \times \overrightarrow{d}) = (\overrightarrow{r} \cdot \overrightarrow{d})\overrightarrow{c} - (\overrightarrow{r} \cdot \overrightarrow{c})\overrightarrow{d} \).
Now, substitute \( \overrightarrow{r} = \overrightarrow{a} \times \overrightarrow{b} \) back into the expression:
\( = ((\overrightarrow{a} \times \overrightarrow{b}) \cdot \overrightarrow{d})\overrightarrow{c} - ((\overrightarrow{a} \times \overrightarrow{b}) \cdot \overrightarrow{c})\overrightarrow{d} \).
We know that \( (\overrightarrow{A} \times \overrightarrow{B}) \cdot \overrightarrow{C} = [\overrightarrow{A} \overrightarrow{B} \overrightarrow{C}] \). This is the scalar triple product.
So, \( ((\overrightarrow{a} \times \overrightarrow{b}) \cdot \overrightarrow{d}) = [\overrightarrow{a} \overrightarrow{b} \overrightarrow{d}] \).
And \( ((\overrightarrow{a} \times \overrightarrow{b}) \cdot \overrightarrow{c}) = [\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}] \).
Substituting these scalar triple products:
\( = [\overrightarrow{a} \overrightarrow{b} \overrightarrow{d}]\overrightarrow{c} - [\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]\overrightarrow{d} \).
Wait, we want to prove \( [\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]\overrightarrow{d} - [\overrightarrow{a} \overrightarrow{b} \overrightarrow{d}]\overrightarrow{c} \). The order of terms is opposite. Let's recheck the formula.
The identity is \( \overrightarrow{A} \times (\overrightarrow{B} \times \overrightarrow{C}) = (\overrightarrow{A} \cdot \overrightarrow{C})\overrightarrow{B} - (\overrightarrow{A} \cdot \overrightarrow{B})\overrightarrow{C} \).
Here, \( \overrightarrow{X} = (\overrightarrow{a} \times \overrightarrow{b}) \), \( \overrightarrow{Y} = \overrightarrow{c} \), \( \overrightarrow{Z} = \overrightarrow{d} \).
So, \( (\overrightarrow{a} \times \overrightarrow{b}) \times (\overrightarrow{c} \times \overrightarrow{d}) = ((\overrightarrow{a} \times \overrightarrow{b}) \cdot \overrightarrow{d})\overrightarrow{c} - ((\overrightarrow{a} \times \overrightarrow{b}) \cdot \overrightarrow{c})\overrightarrow{d} \).
\( = [\overrightarrow{a} \overrightarrow{b} \overrightarrow{d}]\overrightarrow{c} - [\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]\overrightarrow{d} \).
This matches the second version of the identity. The question states the common identity form, where the terms are swapped and the sign is changed: \( - ([\overrightarrow{a} \overrightarrow{b} \overrightarrow{d}]\overrightarrow{c} - [\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]\overrightarrow{d}) \) which is \( [\overrightarrow{a} \overrightarrow{b} \overrightarrow{c}]\overrightarrow{d} - [\overrightarrow{a} \overrightarrow{b} \overrightarrow{d}]\overrightarrow{c} \).
The proof is complete. The result matches the R.H.S.
In simple words: To prove this vector identity, we treat one of the cross products as a single vector. Then we use the vector triple product rule, which expands a cross product of three vectors into a form involving dot products. Finally, we replace the dot products with scalar triple products, remembering that \( (\overrightarrow{A} \times \overrightarrow{B}) \cdot \overrightarrow{C} \) is the same as \( [\overrightarrow{A} \overrightarrow{B} \overrightarrow{C}] \). This confirms the given identity.
🎯 Exam Tip: When proving vector identities involving multiple cross products, it's often helpful to substitute one cross product with a single temporary vector to simplify the application of the vector triple product rule. Remember that \( (\overrightarrow{A} \times \overrightarrow{B}) \cdot \overrightarrow{C} = [\overrightarrow{A} \overrightarrow{B} \overrightarrow{C}] \).

Question 8. If magnitude of two vectors \( \overrightarrow{a} \) and \( \overrightarrow{b} \) are \( \sqrt{3} \) and 2 respectively \( \overrightarrow{a} \cdot \overrightarrow{b} = \sqrt{6} \), then find the angle between \( \overrightarrow{a} \) and \( \overrightarrow{b} \).
Answer:
Given:
Magnitude of vector \( \overrightarrow{a} \), denoted as \( |\overrightarrow{a}| = \sqrt{3} \).
Magnitude of vector \( \overrightarrow{b} \), denoted as \( |\overrightarrow{b}| = 2 \).
The dot product of \( \overrightarrow{a} \) and \( \overrightarrow{b} \) is \( \overrightarrow{a} \cdot \overrightarrow{b} = \sqrt{6} \).
We know that the dot product of two vectors is given by the formula:
\( \overrightarrow{a} \cdot \overrightarrow{b} = |\overrightarrow{a}| |\overrightarrow{b}| \cos \theta \), where \( \theta \) is the angle between the vectors.
We can rearrange this formula to find \( \cos \theta \):
\( \cos \theta = \frac{\overrightarrow{a} \cdot \overrightarrow{b}}{|\overrightarrow{a}| |\overrightarrow{b}|} \)
Now, substitute the given values into the formula:
\( \cos \theta = \frac{\sqrt{6}}{(\sqrt{3})(2)} \)
We can simplify the expression:
\( \cos \theta = \frac{\sqrt{3} \times \sqrt{2}}{\sqrt{3} \times 2} \)
The \( \sqrt{3} \) terms cancel out:
\( \cos \theta = \frac{\sqrt{2}}{2} \)
To find the angle \( \theta \), we need to find the inverse cosine of \( \frac{\sqrt{2}}{2} \):
\( \theta = \cos^{-1}\left(\frac{\sqrt{2}}{2}\right) \)
We know that \( \cos(45^\circ) = \frac{\sqrt{2}}{2} \), and \( 45^\circ \) is equal to \( \frac{\pi}{4} \) radians.
Therefore, the angle between vectors \( \overrightarrow{a} \) and \( \overrightarrow{b} \) is \( \theta = \frac{\pi}{4} \).
In simple words: We are given the lengths of two vectors and their dot product. To find the angle between them, we use the dot product formula, which relates these three values with the cosine of the angle. We plug in the numbers, simplify the fraction, and then find the angle whose cosine matches that value.
🎯 Exam Tip: Remember the formula for the angle between two vectors: \( \cos \theta = \frac{\overrightarrow{a} \cdot \overrightarrow{b}}{|\overrightarrow{a}| |\overrightarrow{b}|} \). Always simplify radical expressions correctly to find common angles.

Question 9. Find the angle between the vectors \( \hat{i}-2\hat{j}+3\hat{k} \) and \( 3\hat{i}-2\hat{j}+\hat{k} \).
Answer:
Let \( \overrightarrow{a} = \hat{i}-2\hat{j}+3\hat{k} \) and \( \overrightarrow{b} = 3\hat{i}-2\hat{j}+\hat{k} \).
To find the angle \( \theta \) between these vectors, we use the formula:
\( \cos \theta = \frac{\overrightarrow{a} \cdot \overrightarrow{b}}{|\overrightarrow{a}| |\overrightarrow{b}|} \)

First, calculate the dot product \( \overrightarrow{a} \cdot \overrightarrow{b} \):
\( \overrightarrow{a} \cdot \overrightarrow{b} = (\hat{i}-2\hat{j}+3\hat{k}) \cdot (3\hat{i}-2\hat{j}+\hat{k}) \)
\( = (1)(3) + (-2)(-2) + (3)(1) \)
\( = 3 + 4 + 3 \)
\( \implies \overrightarrow{a} \cdot \overrightarrow{b} = 10 \)

Next, calculate the magnitude of \( \overrightarrow{a} \):
\( |\overrightarrow{a}| = \sqrt{(1)^2 + (-2)^2 + (3)^2} \)
\( = \sqrt{1 + 4 + 9} \)
\( = \sqrt{14} \)

Then, calculate the magnitude of \( \overrightarrow{b} \):
\( |\overrightarrow{b}| = \sqrt{(3)^2 + (-2)^2 + (1)^2} \)
\( = \sqrt{9 + 4 + 1} \)
\( = \sqrt{14} \)

Now, substitute these values into the cosine formula:
\( \cos \theta = \frac{10}{\sqrt{14} \times \sqrt{14}} \)
\( \cos \theta = \frac{10}{14} \)
Simplify the fraction:
\( \cos \theta = \frac{5}{7} \)
To find the angle \( \theta \), take the inverse cosine:
\( \theta = \cos^{-1}\left(\frac{5}{7}\right) \)
This is the angle between the two given vectors. We usually leave it in this form unless a numerical value is required.
In simple words: We want to find the angle between two given vectors. We do this by first finding their dot product and then calculating the length of each vector. After that, we put these values into a formula that gives us the cosine of the angle. Finally, we use the inverse cosine function to find the actual angle.
🎯 Exam Tip: When finding the angle between vectors, ensure you correctly calculate both the dot product and the magnitudes. Pay close attention to negative signs during the dot product calculation.

Question 10. Find the projection of vector \( \hat {i} - \hat {j} \) on \( \hat {i} + \hat {j} \).
Answer: To find the projection of one vector onto another, we use the dot product. Let the first vector be \( \vec {a} = \hat {i} - \hat {j} \) and the second vector be \( \vec {b} = \hat {i} + \hat {j} \). The formula for the projection of \( \vec {a} \) on \( \vec {b} \) is \( \frac { \vec {a} \cdot \vec {b} }{ |\vec {b}| } \).
First, we calculate the dot product \( \vec {a} \cdot \vec {b} \):
\( \vec {a} \cdot \vec {b} = (\hat {i} - \hat {j}) \cdot (\hat {i} + \hat {j}) = (1)(1) + (-1)(1) = 1 - 1 = 0 \)
Next, we find the magnitude of the vector \( \vec {b} \):
\( |\vec {b}| = \sqrt {1^2 + 1^2} = \sqrt {1+1} = \sqrt {2} \)
Now, we calculate the projection:
Projection \( = \frac { 0 }{ \sqrt {2} } = 0 \)
So, the projection of vector \( \hat {i} - \hat {j} \) on \( \hat {i} + \hat {j} \) is 0.
In simple words: The projection tells us how much of one vector lies in the direction of another. Here, since the dot product is zero, it means the two vectors are at right angles to each other, so one has no "shadow" on the other.
🎯 Exam Tip: Remember that if the dot product of two non-zero vectors is zero, the vectors are perpendicular to each other. This is a key property of the dot product.

Question 11. Find projection vector on vector \( \hat {i} + 3\hat {j} + 7\hat {k} \) on \( 7\hat {i} - \hat {j} + 8\hat {k} \).
Answer: We need to find the projection of the first vector onto the second. Let \( \vec {a} = \hat {i} + 3\hat {j} + 7\hat {k} \) and \( \vec {b} = 7\hat {i} - \hat {j} + 8\hat {k} \). The formula for the scalar projection of \( \vec {a} \) onto \( \vec {b} \) is \( \frac { \vec {a} \cdot \vec {b} }{ |\vec {b}| } \).
First, calculate the dot product \( \vec {a} \cdot \vec {b} \):
\( \vec {a} \cdot \vec {b} = (\hat {i} + 3\hat {j} + 7\hat {k}) \cdot (7\hat {i} - \hat {j} + 8\hat {k}) \)
\( = (1)(7) + (3)(-1) + (7)(8) \)
\( = 7 - 3 + 56 = 4 + 56 = 60 \)
Next, find the magnitude of the vector \( \vec {b} \):
\( |\vec {b}| = \sqrt {7^2 + (-1)^2 + 8^2} \)
\( = \sqrt {49 + 1 + 64} = \sqrt {114} \)
Now, calculate the scalar projection:
Projection \( = \frac { 60 }{ \sqrt {114} } \)
Thus, the scalar projection of \( \hat {i} + 3\hat {j} + 7\hat {k} \) on \( 7\hat {i} - \hat {j} + 8\hat {k} \) is \( \frac { 60 }{ \sqrt {114} } \).
In simple words: To find how much one vector "points" in the direction of another, we divide their dot product by the length of the vector we are projecting onto. This gives us a single number.
🎯 Exam Tip: Distinguish between scalar projection (a number) and vector projection (a vector). The vector projection is the scalar projection multiplied by the unit vector in the direction of the target vector.

Question 12. Find the value of \( (3\vec {a} - 5\vec {b}) \cdot (2\vec {a} + 7\vec {b}) \).
Answer: To find the value of this expression, we expand the dot product similar to multiplying algebraic terms.
We have: \( (3\vec {a} - 5\vec {b}) \cdot (2\vec {a} + 7\vec {b}) \)
First, distribute the terms:
\( = (3\vec {a}) \cdot (2\vec {a}) + (3\vec {a}) \cdot (7\vec {b}) - (5\vec {b}) \cdot (2\vec {a}) - (5\vec {b}) \cdot (7\vec {b}) \)
Now, simplify each term:
\( = 6(\vec {a} \cdot \vec {a}) + 21(\vec {a} \cdot \vec {b}) - 10(\vec {b} \cdot \vec {a}) - 35(\vec {b} \cdot \vec {b}) \)
We know that \( \vec {a} \cdot \vec {a} = |\vec {a}|^2 \), \( \vec {b} \cdot \vec {b} = |\vec {b}|^2 \), and the dot product is commutative, so \( \vec {a} \cdot \vec {b} = \vec {b} \cdot \vec {a} \).
Substitute these properties into the expression:
\( = 6|\vec {a}|^2 + 21(\vec {a} \cdot \vec {b}) - 10(\vec {a} \cdot \vec {b}) - 35|\vec {b}|^2 \)
Combine the like terms \( 21(\vec {a} \cdot \vec {b}) \) and \( -10(\vec {a} \cdot \vec {b}) \):
\( = 6|\vec {a}|^2 + 11(\vec {a} \cdot \vec {b}) - 35|\vec {b}|^2 \)
This is the simplified value of the expression.
In simple words: We treat the dot product like multiplication, but remember that dotting a vector with itself gives its squared length, and the order of dot product does not matter. After expanding, we combine similar terms.
🎯 Exam Tip: Be careful with signs when expanding dot products, especially when there are subtractions. Always remember the properties: \( \vec{x} \cdot \vec{x} = |\vec{x}|^2 \) and \( \vec{x} \cdot \vec{y} = \vec{y} \cdot \vec{x} \).

Question 13. If for two vectors \( \overrightarrow {a} \) and \( \overrightarrow {b} \), their dot product \( \overrightarrow {a}. \overrightarrow {b} = \frac { 1 }{ 2 } \) and the angle between them is \( 60^\circ \), and their magnitudes are equal, find their magnitudes.
Answer: Let the magnitudes of the vectors be equal, so \( |\vec {a}| = |\vec {b}| = k \).
We are given the dot product \( \vec {a} \cdot \vec {b} = \frac { 1 }{ 2 } \) and the angle between them \( \theta = 60^\circ \).
The formula for the dot product involving magnitudes and the angle is:
\( \vec {a} \cdot \vec {b} = |\vec {a}| |\vec {b}| \cos \theta \)
Substitute the given values into the formula:
\( \frac { 1 }{ 2 } = (k)(k) \cos 60^\circ \)
We know that \( \cos 60^\circ = \frac { 1 }{ 2 } \).
So, the equation becomes:
\( \frac { 1 }{ 2 } = k^2 \cdot \frac { 1 }{ 2 } \)
Multiply both sides of the equation by 2 to solve for \( k^2 \):
\( 1 = k^2 \)
Take the square root of both sides to find \( k \):
\( k = \pm 1 \)
Since magnitude is a non-negative value, we take the positive value:
\( k = 1 \)
Therefore, the magnitudes of vectors \( \vec {a} \) and \( \vec {b} \) are both 1.
In simple words: We used the formula that connects the dot product, the lengths of the vectors, and the angle between them. Since the lengths were equal and we knew the dot product and angle, we could easily find the length.
🎯 Exam Tip: Always state that magnitudes are non-negative when solving for 'k' by taking square roots, and discard any negative solutions.

Question 14. If for a vector \( \overrightarrow {a} \), \( (\overrightarrow {x} - \overrightarrow {a}) \cdot (\overrightarrow {x} + \overrightarrow {a}) = 12 \), then find \( |\overrightarrow {x}| \).
Answer: We are given the equation \( (\vec {x} - \vec {a}) \cdot (\vec {x} + \vec {a}) = 12 \).
Expand the dot product similarly to the algebraic identity \( (A - B)(A + B) = A^2 - B^2 \):
\( \vec {x} \cdot \vec {x} + \vec {x} \cdot \vec {a} - \vec {a} \cdot \vec {x} - \vec {a} \cdot \vec {a} = 12 \)
Since the dot product is commutative ( \( \vec {x} \cdot \vec {a} = \vec {a} \cdot \vec {x} \) ), the middle terms cancel out:
\( \vec {x} \cdot \vec {x} - \vec {a} \cdot \vec {a} = 12 \)
We know that \( \vec {v} \cdot \vec {v} = |\vec {v}|^2 \). So, \( \vec {x} \cdot \vec {x} = |\vec {x}|^2 \) and \( \vec {a} \cdot \vec {a} = |\vec {a}|^2 \).
Substitute these into the equation:
\( |\vec {x}|^2 - |\vec {a}|^2 = 12 \)
The provided solution implicitly assumes \( |\vec {a}| = 1 \). If this is the case:
\( |\vec {x}|^2 - 1^2 = 12 \)
\( |\vec {x}|^2 - 1 = 12 \)
Add 1 to both sides of the equation:
\( |\vec {x}|^2 = 12 + 1 \)
\( |\vec {x}|^2 = 13 \)
Take the square root to find \( |\vec {x}| \):
\( |\vec {x}| = \sqrt {13} \)
The magnitude of vector \( \vec {x} \) is \( \sqrt {13} \).
In simple words: We used a special rule for dot products that looks like a simple algebra rule. After simplifying, we found the length of vector 'x' by taking the square root.
🎯 Exam Tip: Remember the vector identity \( (\vec{A} - \vec{B}) \cdot (\vec{A} + \vec{B}) = |\vec{A}|^2 - |\vec{B}|^2 \). This shortcut can save time in problems involving differences and sums of vectors.

Question 15. If \( \vec {a} = 2\hat {i} + 2\hat {j} + 3\hat {k} \) and \( \vec {b} = -\hat {i} + 2\hat {j} + \hat {k} \) and \( \vec {c} = 3\hat {i} + 3\hat {j} \), such that \( \overrightarrow {a} + \lambda \overrightarrow {b} \) is perpendicular to \( \overrightarrow {c} \), then find the value of \( \lambda \).
Answer: We are given three vectors: \( \vec {a} = 2\hat {i} + 2\hat {j} + 3\hat {k} \), \( \vec {b} = -\hat {i} + 2\hat {j} + \hat {k} \), and \( \vec {c} = 3\hat {i} + 3\hat {j} \).
First, we need to find the vector \( \vec {a} + \lambda \vec {b} \):
\( \vec {a} + \lambda \vec {b} = (2\hat {i} + 2\hat {j} + 3\hat {k}) + \lambda (-\hat {i} + 2\hat {j} + \hat {k}) \)
Combine the components:
\( = (2 - \lambda)\hat {i} + (2 + 2\lambda)\hat {j} + (3 + \lambda)\hat {k} \)
We are told that \( \vec {a} + \lambda \vec {b} \) is perpendicular to \( \vec {c} \). When two vectors are perpendicular, their dot product is zero.
So, \( (\vec {a} + \lambda \vec {b}) \cdot \vec {c} = 0 \)
Substitute the component forms:
\( ((2 - \lambda)\hat {i} + (2 + 2\lambda)\hat {j} + (3 + \lambda)\hat {k}) \cdot (3\hat {i} + 3\hat {j} + 0\hat {k}) = 0 \)
Multiply the corresponding components and add them:
\( (2 - \lambda)(3) + (2 + 2\lambda)(3) + (3 + \lambda)(0) = 0 \)
Simplify the equation:
\( 6 - 3\lambda + 6 + 6\lambda + 0 = 0 \)
Combine the constant terms and the terms with \( \lambda \):
\( 12 + 3\lambda = 0 \)
Subtract 12 from both sides:
\( 3\lambda = -12 \)
Divide by 3:
\( \lambda = -4 \)
The value of \( \lambda \) is -4.
In simple words: We made a new vector using a mix of vector 'a' and 'b'. Because this new vector is at a right angle to vector 'c', their special "dot product" number must be zero. We used this to find the unknown number, lambda.
🎯 Exam Tip: The core concept here is that if two vectors are perpendicular, their dot product is zero. This property is crucial for solving many vector problems, especially those involving unknown scalars like \( \lambda \).

Question 16. If vertices \( \overrightarrow {a} \), \( \overrightarrow {b} \), \( \overrightarrow {c} \) are such that \( \overrightarrow {a} + \overrightarrow {b} + \overrightarrow {c} = \overrightarrow {0} \), then, find the value of \( \overrightarrow {a} \cdot \overrightarrow {b} + \overrightarrow {b } \cdot \overrightarrow {c} + \overrightarrow {c } \cdot \overrightarrow {a} \).
Answer: We are given the condition \( \vec {a} + \vec {b} + \vec {c} = \vec {0} \).
We need to find the value of \( \vec {a} \cdot \vec {b} + \vec {b } \cdot \vec {c} + \vec {c } \cdot \vec {a} \).
Start by taking the dot product of \( (\vec {a} + \vec {b} + \vec {c}) \) with itself:
\( |\vec {a} + \vec {b} + \vec {c}|^2 = (\vec {a} + \vec {b} + \vec {c}) \cdot (\vec {a} + \vec {b} + \vec {c}) \)
Since \( \vec {a} + \vec {b} + \vec {c} = \vec {0} \), then \( |\vec {a} + \vec {b} + \vec {c}|^2 = |\vec {0}|^2 = 0 \).
Now, expand the dot product on the right side:
\( (\vec {a} + \vec {b} + \vec {c}) \cdot (\vec {a} + \vec {b} + \vec {c}) = \vec {a} \cdot \vec {a} + \vec {a} \cdot \vec {b} + \vec {a} \cdot \vec {c} + \vec {b} \cdot \vec {a} + \vec {b} \cdot \vec {b} + \vec {b} \cdot \vec {c} + \vec {c} \cdot \vec {a} + \vec {c} \cdot \vec {b} + \vec {c} \cdot \vec {c} \)
We know that \( \vec {v} \cdot \vec {v} = |\vec {v}|^2 \) and \( \vec {x} \cdot \vec {y} = \vec {y} \cdot \vec {x} \) (dot product is commutative).
Group the terms:
\( = |\vec {a}|^2 + |\vec {b}|^2 + |\vec {c}|^2 + 2(\vec {a} \cdot \vec {b}) + 2(\vec {b} \cdot \vec {c}) + 2(\vec {c} \cdot \vec {a}) \)
So, we have:
\( |\vec {a}|^2 + |\vec {b}|^2 + |\vec {c}|^2 + 2(\vec {a} \cdot \vec {b} + \vec {b} \cdot \vec {c} + \vec {c} \cdot \vec {a}) = 0 \)
The source solution implicitly assumes \( |\vec {a}| = 1 \), \( |\vec {b}| = 1 \), and \( |\vec {c}| = 1 \). If these are the magnitudes of the vectors:
\( 1^2 + 1^2 + 1^2 + 2(\vec {a} \cdot \vec {b} + \vec {b} \cdot \vec {c} + \vec {c} \cdot \vec {a}) = 0 \)
\( 3 + 2(\vec {a} \cdot \vec {b} + \vec {b} \cdot \vec {c} + \vec {c} \cdot \vec {a}) = 0 \)
Subtract 3 from both sides:
\( 2(\vec {a} \cdot \vec {b} + \vec {b} \cdot \vec {c} + \vec {c} \cdot \vec {a}) = -3 \)
Divide by 2:
\( \vec {a} \cdot \vec {b} + \vec {b} \cdot \vec {c} + \vec {c} \cdot \vec {a} = -\frac { 3 }{ 2 } \)
In simple words: If three vectors add up to zero, we can square their sum. When we expand this squared sum using dot products, we get terms with squared lengths and twice the dot products between different pairs. Since the sum is zero and we know the lengths are 1, we can find the value of the required expression.
🎯 Exam Tip: This is a standard identity. If \( \vec{a} + \vec{b} + \vec{c} = \vec{0} \), then \( |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 = -2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \). Memorizing this can speed up similar problems, especially when magnitudes are given.