Anand Classes provides clear and well-structured NCERT Solutions for Exercise 10.4 of Chapter 10 Vector Algebra for Class 12 Mathematics, helping students master the concept of the cross product, its properties, and applications through detailed, step-by-step explanations. These Set-1 notes strictly follow the latest NCERT and CBSE syllabus, making them highly useful for revision, concept clarity, and effective preparation for board exams. Click the print button to download study material and notes.
NCERT Question.1 : Find $|\vec{a}\times \vec{b}|$ if
$\vec{a}= \hat{i}-7\hat{j}+7\hat{k}$ and $\vec{b}=3\hat{i}-2\hat{j}+2\hat{k}$.
Solution
The given vectors are
$$
\vec{a}= \hat{i}-7\hat{j}+7\hat{k}
$$
$$
\vec{b}= 3\hat{i}-2\hat{j}+2\hat{k}
$$
Step 1: Compute the cross product $\vec{a}\times \vec{b}$
$$
\vec{a}\times \vec{b}=
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
1 & -7 & 7 \\
3 & -2 & 2
\end{vmatrix}
$$
$$
\vec{a}\times \vec{b}=
\hat{i}((-7)(2)-(7)(-2))
-\hat{j}((1)(2)-(7)(3))
+\hat{k}((1)(-2)-(-7)(3))
$$
$$
\vec{a}\times \vec{b}=
\hat{i}(-14+14)
-\hat{j}(2-21)
+\hat{k}(-2+21)
$$
$$
\vec{a}\times \vec{b}=0\hat{i}+19\hat{j}+19\hat{k}
$$
Step 2: Magnitude of the cross product
$$
|\vec{a}\times \vec{b}|=\sqrt{0^{2}+19^{2}+19^{2}}
$$
$$
|\vec{a}\times \vec{b}|=\sqrt{361+361}
$$
$$
|\vec{a}\times \vec{b}|=19\sqrt{2}
$$
Final Result
$$
\boxed{19\sqrt{2}}
$$
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NCERT Question.2 : Find a unit vector perpendicular to each of the vectors $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$, where
$\vec{a}=3\hat{i}+2\hat{j}+2\hat{k}$ and $\vec{b}=\hat{i}+2\hat{j}-2\hat{k}$.
Solution
Step 1: Compute $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$
$$
\vec{a}+\vec{b}=(3\hat{i}+2\hat{j}+2\hat{k})+(\hat{i}+2\hat{j}-2\hat{k})
$$
$$
\vec{a}+\vec{b}=4\hat{i}+4\hat{j}+0\hat{k}
$$
$$
\vec{a}-\vec{b}=(3\hat{i}+2\hat{j}+2\hat{k})-(\hat{i}+2\hat{j}-2\hat{k})
$$
$$
\vec{a}-\vec{b}=2\hat{i}+0\hat{j}+4\hat{k}
$$
Step 2: Cross product $(\vec{a}+\vec{b})\times(\vec{a}-\vec{b})$
$$
(\vec{a}+\vec{b})\times(\vec{a}-\vec{b})=
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
4 & 4 & 0 \\
2 & 0 & 4
\end{vmatrix}
$$
$$ (\vec{a}+\vec{b})\times(\vec{a}-\vec{b})= \hat{i}(4\cdot 4 – 0\cdot 0) -\hat{j}(4\cdot 4 – 0\cdot 2) +\hat{k}(4\cdot 0 – 4\cdot 2) $$
$$
(\vec{a}+\vec{b})\times(\vec{a}-\vec{b})= 16\hat{i} – 16\hat{j} – 8\hat{k}
$$
Let
$$
\vec{c}=16\hat{i}-16\hat{j}-8\hat{k}
$$
Step 3: Magnitude of $\vec{c}$
$$
|\vec{c}|=\sqrt{16^{2}+(-16)^{2}+(-8)^{2}}
$$
$$
|\vec{c}|=\sqrt{256+256+64}
$$
$$
|\vec{c}|=24
$$
Step 4: Unit vector perpendicular to both vectors
$$
\hat{c}=\pm\frac{\vec{c}}{|\vec{c}|}
=\pm\frac{16\hat{i}-16\hat{j}-8\hat{k}}{24}
$$
Factor $8$:
$$
\hat{c}=\pm\frac{2\hat{i}-2\hat{j}-\hat{k}}{3}
$$
Final Result
$$
\boxed{\pm\left(\frac{2}{3}\hat{i}-\frac{2}{3}\hat{j}-\frac{1}{3}\hat{k}\right)}
$$
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NCERT Question.3 : If a unit vector $\vec{a}$ makes angles $\frac{\pi}{3}$ with $\hat{i}$, $\frac{\pi}{4}$ with $\hat{j}$ and an acute angle $\theta$ with $\hat{k}$, then find $\theta$ and hence the components of $\vec{a}$.
Solution
For a vector making angles $\alpha$, $\beta$, $\gamma$ with the coordinate axes, the direction cosines satisfy
$$\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma = 1$$
Given:
- $\alpha = \frac{\pi}{3}$
- $\beta = \frac{\pi}{4}$
- $\gamma = \theta$ (acute)
Substitute the values:
$$\cos^2 \left( \frac{\pi}{3} \right) + \cos^2 \left( \frac{\pi}{4} \right) + \cos^2 \theta = 1$$
Compute the cosine values:
$$\cos \left( \frac{\pi}{3} \right) = \frac{1}{2} \Rightarrow \cos^2 \left( \frac{\pi}{3} \right) = \frac{1}{4}$$
$$\cos \left( \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}} \Rightarrow \cos^2 \left( \frac{\pi}{4} \right) = \frac{1}{2}$$
Substitute them:
$$\frac{1}{4} + \frac{1}{2} + \cos^2 \theta = 1$$
$$\frac{3}{4} + \cos^2 \theta = 1$$
$$\cos^2 \theta = \frac{1}{4}$$
Since $\theta$ is acute:
$$\cos \theta = \frac{1}{2}$$
Thus:
$$\theta = \frac{\pi}{3}$$
Components of the Unit Vector
A unit vector $\vec{a}$ having direction cosines $l, m, n$ is written as
$$\vec{a} = l \hat{i} + m \hat{j} + n \hat{k}$$
Compute each:
$$l = \cos \alpha = \cos \left( \frac{\pi}{3} \right) = \frac{1}{2}$$
$$m = \cos \beta = \cos \left( \frac{\pi}{4} \right) = \frac{1}{\sqrt{2}}$$
$$n = \cos \theta = \cos \left( \frac{\pi}{3} \right) = \frac{1}{2}$$
Thus the required unit vector is
$$\vec{a} = \frac{1}{2} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} + \frac{1}{2} \hat{k}$$
Final Result
$$\boxed{\theta = \frac{\pi}{3} \quad \text{and} \quad \vec{a} = \frac{1}{2} \hat{i} + \frac{1}{\sqrt{2}} \hat{j} + \frac{1}{2} \hat{k}}$$
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NCERT Question.4 : Show that
$$(\vec{a}-\vec{b}) \times (\vec{a}+\vec{b}) = 2(\vec{a} \times \vec{b})$$
Solution
Start with the given expression:
$$ (\vec{a}-\vec{b}) \times (\vec{a}+\vec{b}) $$
Apply the distributive property of the cross product:
$$ (\vec{a}-\vec{b}) \times (\vec{a}+\vec{b}) = \vec{a} \times (\vec{a}+\vec{b}) – \vec{b} \times (\vec{a}+\vec{b}) $$
Expand each term:
$$ (\vec{a}-\vec{b}) \times (\vec{a}+\vec{b})= (\vec{a} \times \vec{a}) + (\vec{a} \times \vec{b}) – (\vec{b} \times \vec{a}) – (\vec{b} \times \vec{b}) $$
Use the identities:
- $\vec{a} \times \vec{a} = \vec{0}$
- $\vec{b} \times \vec{b} = \vec{0}$
- $\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b})$
Substitute:
$$ (\vec{a}-\vec{b}) \times (\vec{a}+\vec{b})= \vec{0} + (\vec{a} \times \vec{b}) – \left( -(\vec{a} \times \vec{b}) \right) – \vec{0} $$
Simplify:
$$ (\vec{a}-\vec{b}) \times (\vec{a}+\vec{b})= \vec{a} \times \vec{b} + \vec{a} \times \vec{b} $$
$$ (\vec{a}-\vec{b}) \times (\vec{a}+\vec{b})= 2(\vec{a} \times \vec{b}) $$
Final Result
$$\boxed{(\vec{a}-\vec{b}) \times (\vec{a}+\vec{b}) = 2(\vec{a} \times \vec{b})}$$
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NCERT Question.5 : Find $\lambda$ and $\mu$ if
$$(2\hat{i}+6\hat{j}+27\hat{k}) \times (\hat{i}+\lambda \hat{j}+\mu \hat{k}) = \vec{0}$$
Solution
Let
$$\vec{a} = 2\hat{i} + 6\hat{j} + 27\hat{k}$$
$$\vec{b} = \hat{i} + \lambda \hat{j} + \mu \hat{k}$$
Compute the cross product using the determinant:
$$
\vec{a} \times \vec{b} =
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
2 & 6 & 27 \\
1 & \lambda & \mu
\end{vmatrix}
$$
Expand along the first row:
$$ \vec{a} \times \vec{b} = \hat{i}(6\mu – 27\lambda)-\hat{j}(2\mu – 27)+\hat{k}(2\lambda – 6)$$
Since
$$\vec{a} \times \vec{b} = \vec{0} = 0\hat{i}+0\hat{j}+0\hat{k}$$
each component must be zero.
From the $\hat{k}$–component:
$$2\lambda – 6 = 0$$
$$2\lambda = 6$$
$$\lambda = 3$$
From the $\hat{j}$–component:
$$-(2\mu – 27) = 0$$
$$2\mu – 27 = 0$$
$$2\mu = 27$$
$$\mu = \frac{27}{2}$$
Verification using the $\hat{i}$–component:
$$6\mu – 27\lambda = 0$$
Substitute $\lambda = 3$ and $\mu = \dfrac{27}{2}$:
$$6 \left(\frac{27}{2}\right) – 27(3) = 0$$
$$81 – 81 = 0$$
True ✔️
Final Result
$$\boxed{\lambda = 3 \quad,\quad \mu = \frac{27}{2}}$$
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NCERT Question.6 : Given that
$$\vec{a}\cdot \vec{b} = 0 \quad \text{and} \quad \vec{a}\times \vec{b} = \vec{0}$$
What can you conclude about the vectors $\vec{a}$ and $\vec{b}$?
Solution
Condition 1: Dot product is zero
$$\vec{a}\cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta = 0$$
This is possible only if:
- $|\vec{a}| = 0$ (i.e., $\vec{a}$ is the zero vector), or
- $|\vec{b}| = 0$ (i.e., $\vec{b}$ is the zero vector), or
- $\cos\theta = 0$ which implies $$\theta = \frac{\pi}{2}$$ (vectors are perpendicular)
Condition 2: Cross product is zero
$$|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin\theta = 0$$
This is possible only if:
- $|\vec{a}| = 0$
- $|\vec{b}| = 0$
- $\sin\theta = 0$ which implies $$\theta = 0 \quad \text{or} \quad \theta = \pi$$
(vectors are parallel or anti-parallel)
Combine the Conditions
Both must hold simultaneously.
If $\vec{a}$ and $\vec{b}$ are non-zero, then:
- From dot product: $\theta = \frac{\pi}{2}$
- From cross product: $\theta = 0$ or $\pi$
A single angle cannot be both $\frac{\pi}{2}$ and $0$ or $\pi$.
This is impossible unless one or both vectors are zero.
Final Result
$$\boxed{\text{At least one of the vectors } \vec{a} \text{ or } \vec{b} \text{ must be the zero vector.}}$$
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