Anand Classes offers complete NCERT Solutions for Exercise 10.4 of Chapter 10 Vector Algebra for Class 12 Mathematics, helping students understand the cross product of vectors, its geometric interpretation, and related problem-solving techniques through clear, step-by-step solutions. These Set-2 notes are based on the latest NCERT and CBSE curriculum, making them ideal for revision, concept strengthening, and scoring high in board exams. Click the print button to download study material and notes.
NCERT Question.7 : Let the given vectors
$\vec{a} = a_{1}\vec{i} + a_{2} \vec{j} + a_{3} \vec{k}$
$\vec{b} = b_{1} \vec{i} + b_{2} \vec{j} + b_{3} \vec{k}$
$\vec{c} = c_{1} \vec{i} + c_{2} \vec{j} + c_{3} \vec{k}$
Show that
$$\vec{a} \times (\vec{b} + \vec{c}) = (\vec{a} \times \vec{b}) + (\vec{a} \times \vec{c}).$$
Solution
The given vectors are :
$\vec{a} = a_{1}\vec{i} + a_{2} \vec{j} + a_{3} \vec{k}$
$\vec{b} = b_{1} \vec{i} + b_{2} \vec{j} + b_{3} \vec{k}$
$\vec{c} = c_{1} \vec{i} + c_{2} \vec{j} + c_{3} \vec{k}$
1. Compute the Left-Hand Side (LHS)
First add the vectors:
$$\vec{b} + \vec{c} = (b_{1} + c_{1}) \vec{i} + (b_{2} + c_{2}) \vec{j} + (b_{3} + c_{3}) \vec{k}$$
Now compute the cross product:
$$
\vec{a} \times (\vec{b} + \vec{c}) =
\begin{vmatrix}
\vec{i} & \vec{j} & \vec{k} \\
a_{1} & a_{2} & a_{3} \\
(b_{1}+c_{1}) & (b_{2}+c_{2}) & (b_{3}+c_{3})
\end{vmatrix}
$$
Expanding:
$$
\vec{a} \times (\vec{b} + \vec{c})=\vec{i}[a_{2}(b_{3}+c_{3}) – a_{3}(b_{2}+c_{2})]
-\vec{j}[a_{1}(b_{3}+c_{3}) – a_{3}(b_{1}+c_{1})]\\[1em]
+\vec{k}[a_{1}(b_{2}+c_{2}) – a_{2}(b_{1}+c_{1})]
$$
Rewriting terms:
$$
\vec{a} \times (\vec{b} + \vec{c})=\vec{i}(a_{2}b_{3} – a_{3}b_{2} + a_{2}c_{3} – a_{3}c_{2})
-\vec{j}(a_{1}b_{3} – a_{3}b_{1} + a_{1}c_{3} – a_{3}c_{1})\\[1em]
+\vec{k}(a_{1}b_{2} – a_{2}b_{1} + a_{1}c_{2} – a_{2}c_{1})
\quad (1)
$$
2. Compute the Right-Hand Side (RHS)
Compute $(\vec{a} \times \vec{b})$:
$$
\vec{a} \times \vec{b} =
\begin{vmatrix}
\vec{i} & \vec{j} & \vec{k} \\
a_{1} & a_{2} & a_{3} \\
b_{1} & b_{2} & b_{3}
\end{vmatrix}$$
$$\vec{a} \times \vec{b} =\vec{i}(a_{2}b_{3} – a_{3}b_{2})
-\vec{j}(a_{1}b_{3} – a_{3}b_{1})
+\vec{k}(a_{1}b_{2} – a_{2}b_{1})$$
Compute $(\vec{a} \times \vec{c})$:
$$
\vec{a} \times \vec{c} =
\begin{vmatrix}
\vec{i} & \vec{j} & \vec{k} \\
a_{1} & a_{2} & a_{3} \\
c_{1} & c_{2} & c_{3}
\end{vmatrix}$$
$$\vec{a} \times \vec{c} =\vec{i}(a_{2}c_{3} – a_{3}c_{2})
-\vec{j}(a_{1}c_{3} – a_{3}c_{1})
+\vec{k}(a_{1}c_{2} – a_{2}c_{1})
$$
Add the results:
$$ (\vec{a} \times \vec{b}) + (\vec{a} \times \vec{c})=
\vec{i}(a_{2}b_{3} – a_{3}b_{2} + a_{2}c_{3} – a_{3}c_{2})\\[1em]
-\vec{j}(a_{1}b_{3} – a_{3}b_{1} + a_{1}c_{3} – a_{3}c_{1})
+\vec{k}(a_{1}b_{2} – a_{2}b_{1} + a_{1}c_{2} – a_{2}c_{1})
\quad (2)
$$
Final Result
By comparing expressions (1) and (2):
$$\boxed{\vec{a} \times (\vec{b} + \vec{c}) = (\vec{a} \times \vec{b}) + (\vec{a} \times \vec{c})}$$
This confirms the distributive property of the vector cross product.
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NCERT Question.8 : If either $\vec a = \vec 0$ or $\vec b = \vec 0$, then $\vec a \times \vec b = \vec 0$. Is the converse true? Justify your answer with an example.
Solution
The converse of the statement is not true.
The magnitude of the cross product is
$$|\vec a \times \vec b| = |\vec a||\vec b|\sin\theta$$
where $\theta$ is the angle between $\vec a$ and $\vec b$.
For
$$\vec a \times \vec b = \vec 0$$
we must have
$$|\vec a||\vec b|\sin\theta = 0.$$
This happens if:
- $|\vec a| = 0$ meaning $\vec a = \vec 0$
- $|\vec b| = 0$ meaning $\vec b = \vec 0$
- $\sin\theta = 0$ meaning $\theta = 0^\circ$ or $180^\circ$
(vectors are parallel or collinear)
The converse claims:
If $\vec a \times \vec b = \vec 0$, then either $\vec a = \vec 0$ or $\vec b = \vec 0$.
This is false, because the vectors may be non-zero but parallel.
Example
Let
$$\vec a = \hat i + \hat j$$
$$\vec b = 2\hat i + 2\hat j$$
Both vectors are non-zero.
Compute the cross product:
$$\vec a \times \vec b =
\begin{vmatrix}
\hat i & \hat j & \hat k\\
1 & 1 & 0\\
2 & 2 & 0
\end{vmatrix}$$
Expanding:
$$\vec a \times \vec b = \hat i(1\cdot 0 – 0\cdot 2) – \hat j(1\cdot 0 – 0\cdot 2) + \hat k(1\cdot 2 – 1\cdot 2)$$
$$\vec a \times \vec b = \hat i(0) – \hat j(0) + \hat k(0)$$
$$\vec a \times \vec b = \vec 0$$
Here
$$\vec b = 2\vec a$$
so the vectors are collinear, not zero.
Thus, the converse is not true.
NCERT Question.9 : Find the area of the triangle with vertices $A(1,1,2), B(2,3,5)$ and $C(1,5,5)$.
Solution
Compute side vectors
$$\overrightarrow{AB}=\vec{B}-\vec{A}=(2-1)\hat{i}+(3-1)\hat{j}+(5-2)\hat{k}=1\hat{i}+2\hat{j}+3\hat{k}$$
$$\overrightarrow{AC}=\vec{C}-\vec{A}=(1-1)\hat{i}+(5-1)\hat{j}+(5-2)\hat{k}=0\hat{i}+4\hat{j}+3\hat{k}$$
Cross product
$$\overrightarrow{AB}\times\overrightarrow{AC}=
\begin{vmatrix}
\;\hat{i} & \hat{j} & \hat{k} \;\\
\;1 & 2 & 3 \;\\
\;0 & 4 & 3\;
\end{vmatrix}$$
$$\overrightarrow{AB}\times\overrightarrow{AC}=\hat{i}(2\cdot 3-3\cdot 4)-\hat{j}(1\cdot 3-3\cdot 0)+\hat{k}(1\cdot 4-2\cdot 0)$$
$$\overrightarrow{AB}\times\overrightarrow{AC}= -6\hat{i}-3\hat{j}+4\hat{k}$$
Magnitude
$$\bigl|\overrightarrow{AB}\times\overrightarrow{AC}\bigr|=\sqrt{(-6)^2+(-3)^2+4^2}=\sqrt{36+9+16}=\sqrt{61}$$
Area of triangle
$$\text{Area}=\frac{1}{2}\bigl|\overrightarrow{AB}\times\overrightarrow{AC}\bigr|=\frac{\sqrt{61}}{2}$$
Final Result
$$\boxed{\text{Area}=\frac{\sqrt{61}}{2}\ \text{square units}}$$
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NCERT Question.10 : Find the area of the parallelogram whose adjacent sides are determined by the vectors
$\vec{a}=\hat{i}-\hat{j}+3\hat{k}$ and $\vec{b}=2\hat{i}-7\hat{j}+\hat{k}$
Solution
The area of a parallelogram with adjacent sides $\vec{a}$ and $\vec{b}$ is given by the magnitude of their cross product, $|\vec{a}\times\vec{b}|$. First, calculate the cross product:
$$\vec{a}\times\vec{b}=\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k}\\
1 & -1 & 3\\
2 & -7 & 1
\end{vmatrix}$$
Expand
$$\vec{a}\times\vec{b}=\hat{i}(-1\cdot1-3\cdot-7)-\hat{j}(1\cdot1-3\cdot2)+\hat{k}(1\cdot-7-(-1)\cdot2)$$
$$\vec{a}\times\vec{b}=20\hat{i}+5\hat{j}-5\hat{k}$$
Magnitude
$$|\vec{a}\times\vec{b}|=\sqrt{20^2+5^2+(-5)^2}$$
$$|\vec{a}\times\vec{b}|=\sqrt{400+25+25}$$
$$|\vec{a}\times\vec{b}|=\sqrt{450}$$
$$|\vec{a}\times\vec{b}|=15\sqrt{2}$$
Final Result
$$\boxed{15\sqrt{2}}$$
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NCERT Question.11 : Let the vectors $\vec{a}$ and $\vec{b}$ satisfy $|\vec{a}|=3$ and $|\vec{b}|=\frac{\sqrt{2}}{3}$.
If $\vec{a}\times\vec{b}$ is a unit vector, find the angle between $\vec{a}$ and $\vec{b}$.
Solution
Use the magnitude formula
$$|\vec{a}\times\vec{b}|=|\vec{a}||\vec{b}|\sin\theta$$
Given
$$|\vec{a}|=3$$
$$|\vec{b}|=\frac{\sqrt{2}}{3}$$
and
$$|\vec{a}\times\vec{b}|=1$$
Substitute
$$|\vec{a}\times\vec{b}|=|\vec{a}||\vec{b}|\sin\theta$$
$$1=3\cdot\frac{\sqrt{2}}{3}\sin\theta$$
Simplify
$$1=\sqrt{2}\sin\theta$$
$$\sin\theta=\frac{1}{\sqrt{2}}$$
Thus
$$\theta=\frac{\pi}{4}$$
Final Result
$$\boxed{\frac{\pi}{4}}$$
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NCERT Question.12 : Area of a rectangle having vertices $ A, B, C, D $ with position vectors
$\vec{A} = -\hat{i} + \frac{1}{2}\hat{j} + 4\hat{k}\;$, $\vec{B} = \hat{i} + \frac{1}{2}\hat{j} + 4\hat{k}\;$, $\vec{C} = \hat{i} – \frac{1}{2}\hat{j} + 4\hat{k}\;$, $\vec{D} = -\hat{i} – \frac{1}{2}\hat{j} + 4\hat{k}$
is
(A) $\frac{1}{2}$ (B) $1$ (C) $2$ (D) $4$
Solution
Step 1: Write the position vectors
$$\vec{A} = -\hat{i} + \frac{1}{2}\hat{j} + 4\hat{k}$$
$$\vec{B} = \hat{i} + \frac{1}{2}\hat{j} + 4\hat{k}$$
$$\vec{C} = \hat{i} – \frac{1}{2}\hat{j} + 4\hat{k}$$
$$\vec{D} = -\hat{i} – \frac{1}{2}\hat{j} + 4\hat{k}$$
Step 2: Find two adjacent side vectors
$$\overrightarrow{AB} = \vec{B} – \vec{A}$$
$$\overrightarrow{AB}= \left(\hat{i} + \frac{1}{2}\hat{j} + 4\hat{k}\right) – \left(-\hat{i} + \frac{1}{2}\hat{j} + 4\hat{k}\right)$$
$$\overrightarrow{AB}= 2\hat{i}$$
$$\overrightarrow{AD} = \vec{D} – \vec{A}$$
$$\overrightarrow{AD}= \left(-\hat{i} – \frac{1}{2}\hat{j} + 4\hat{k}\right) – \left(-\hat{i} + \frac{1}{2}\hat{j} + 4\hat{k}\right)$$
$$\overrightarrow{AD}= -\hat{j}$$
Step 3: Compute the cross product
$$\overrightarrow{AB} \times \overrightarrow{AD} = (2\hat{i}) \times (-\hat{j})$$
$$\overrightarrow{AB} \times \overrightarrow{AD}= -2(\hat{i} \times \hat{j})$$
$$\overrightarrow{AB} \times \overrightarrow{AD}= -2\hat{k}$$
Step 4: Area of the rectangle
Magnitude of cross product gives the area of rectangle
$$\text{Area} =|\overrightarrow{AB} \times \overrightarrow{AD}|$$
$$\text{Area} = \left| -2\hat{k} \right| = 2$$
Final Result
$$\boxed{2}$$
The correct option is (C).
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