Where Mindful Learning Begins
Where Mindful Learning Begins
This Class 12 Maths NCERT Solutions Chapter 10 Ex 10.4 page covers all 12 questions, solved step-by-step, on the vector (cross) product \vec{a}\times\vec{b}=|\vec{a}||\vec{b}|\sin\theta\,\hat{n} — the operation that, unlike the dot product, hands back another vector: one perpendicular to both of the vectors you started with.
Questions 1–3 build the mechanics: computing a cross product using the \hat{i},\hat{j},\hat{k} determinant, and using it to find a unit vector perpendicular to a plane — the standalone skill Chapter 11 (Three Dimensional Geometry) leans on constantly for finding normal vectors to planes. Questions 4–8 are proof and reasoning questions: identities involving the cross product, the condition for two vectors to be parallel (cross product zero), and — a favourite conceptual trap — what it means when both the dot product and the cross product of two vectors are zero simultaneously. Questions 9–12 turn to the exercise's most exam-relevant application: using \frac12|\vec{AB}\times\vec{AC}| for the area of a triangle and |\vec{a}\times\vec{b}| for the area of a parallelogram, which shows up as a 2- or 3-mark question in almost every CBSE sample paper for this chapter.
\vec{a}\times\vec{b} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&-7&7\\3&-2&2\end{vmatrix}
= \hat{i}[(-7)(2)-(7)(-2)] - \hat{j}[(1)(2)-(7)(3)] + \hat{k}[(1)(-2)-(-7)(3)]
= \hat{i}(-14+14) - \hat{j}(2-21) + \hat{k}(-2+21) = 0\hat{i}+19\hat{j}+19\hat{k}.
|\vec{a}\times\vec{b}| = \sqrt{0+361+361} = \sqrt{722} = 19\sqrt2.
\vec{a}+\vec{b} = 4\hat{i}+4\hat{j}+0\hat{k} and \vec{a}-\vec{b} = 2\hat{i}+0\hat{j}+4\hat{k}.
A vector perpendicular to both is (\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}
= \hat{i}(16-0) - \hat{j}(16-0) + \hat{k}(0-8) = 16\hat{i}-16\hat{j}-8\hat{k} (call this \vec{c}).
|\vec{c}| = \sqrt{256+256+64} = \sqrt{576} = 24.
Required unit vector = \dfrac{\vec{c}}{|\vec{c}|} = \dfrac23\hat{i}-\dfrac23\hat{j}-\dfrac13\hat{k}.
Direction cosines: l=\cos\frac{\pi}{3}=\frac12, m=\cos\frac{\pi}{4}=\frac{1}{\sqrt2}, n=\cos\theta.
Since l^2+m^2+n^2=1: \frac14+\frac12+n^2=1 \Rightarrow n^2=\frac14 \Rightarrow n=\frac12 (taking the positive root since θ is acute).
So \cos\theta=\frac12 \Rightarrow \theta=\frac{\pi}{3}.
Components of \vec{a}: \vec{a}=\frac12\hat{i}+\frac{1}{\sqrt2}\hat{j}+\frac12\hat{k}.
Expanding using distributivity of the cross product over addition:
(\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}.
Using \vec{a}\times\vec{a}=\vec{0}, \vec{b}\times\vec{b}=\vec{0} and \vec{b}\times\vec{a}=-\vec{a}\times\vec{b}:
= \vec{0} + \vec{a}\times\vec{b} - (-\vec{a}\times\vec{b}) - \vec{0} = 2(\vec{a}\times\vec{b}).
The cross product of two nonzero vectors is the zero vector only when the vectors are parallel — that is, their respective components are proportional.
\dfrac{2}{1} = \dfrac{6}{\lambda} = \dfrac{27}{\mu}.
From \frac{2}{1}=\frac{6}{\lambda}: \lambda=3. From \frac{2}{1}=\frac{27}{\mu}: \mu=\dfrac{27}{2}.
\vec{a}\cdot\vec{b}=0 means \vec{a}=\vec{0}, or \vec{b}=\vec{0}, or \vec{a}\perp\vec{b}.
\vec{a}\times\vec{b}=\vec{0} means \vec{a}=\vec{0}, or \vec{b}=\vec{0}, or \vec{a}\parallel\vec{b}.
A pair of nonzero vectors cannot be simultaneously perpendicular and parallel, so both conditions can hold together only if at least one of the vectors is the zero vector.
Here \vec{b}+\vec{c} = (b_1+c_1)\hat{i}+(b_2+c_2)\hat{j}+(b_3+c_3)\hat{k}.
Expanding \vec{a}\times(\vec{b}+\vec{c}) as a determinant and simplifying each component:
\hat{i}\text{-component: } a_2(b_3+c_3)-a_3(b_2+c_2) = (a_2b_3-a_3b_2)+(a_2c_3-a_3c_2)
\hat{j}\text{-component: } -[a_1(b_3+c_3)-a_3(b_1+c_1)] = -(a_1b_3-a_3b_1)-(a_1c_3-a_3c_1)
\hat{k}\text{-component: } a_1(b_2+c_2)-a_2(b_1+c_1) = (a_1b_2-a_2b_1)+(a_1c_2-a_2c_1)
Each component splits cleanly into a "b" part and a "c" part, which are exactly the corresponding components of \vec{a}\times\vec{b} and \vec{a}\times\vec{c}.
No, the converse is not true. Take \vec{a}=\hat{i} and \vec{b}=2\hat{i}, neither of which is the zero vector.
\vec{a}\times\vec{b} = \hat{i}\times2\hat{i} = 2(\hat{i}\times\hat{i}) = 2\vec{0}=\vec{0}, since parallel vectors always have zero cross product.
\vec{AB} = \hat{i}+2\hat{j}+3\hat{k}, \vec{AC} = 0\hat{i}+4\hat{j}+3\hat{k}.
\vec{AB}\times\vec{AC} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&2&3\\0&4&3\end{vmatrix} = \hat{i}(6-12)-\hat{j}(3-0)+\hat{k}(4-0) = -6\hat{i}-3\hat{j}+4\hat{k}.
|\vec{AB}\times\vec{AC}| = \sqrt{36+9+16} = \sqrt{61}.
Area = \dfrac12|\vec{AB}\times\vec{AC}|.
\vec{a}\times\vec{b} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&-1&3\\2&-7&1\end{vmatrix} = \hat{i}(-1+21)-\hat{j}(1-6)+\hat{k}(-7+2)
= 20\hat{i}+5\hat{j}-5\hat{k}.
|\vec{a}\times\vec{b}| = \sqrt{400+25+25} = \sqrt{450} = 15\sqrt2.
\vec{a}\times\vec{b} is a unit vector means |\vec{a}\times\vec{b}|=1, i.e. |\vec{a}||\vec{b}|\sin\theta=1.
3\cdot\dfrac{\sqrt2}{3}\cdot\sin\theta = 1 \Rightarrow \sqrt2\sin\theta=1 \Rightarrow \sin\theta=\dfrac{1}{\sqrt2}.
\vec{AB} = (1-(-1))\hat{i}+0\hat{j}+0\hat{k} = 2\hat{i}.
\vec{AD} = (-1-(-1))\hat{i}+(-\frac12-\frac12)\hat{j}+0\hat{k} = -\hat{j}.
\vec{AB}\times\vec{AD} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&0&0\\0&-1&0\end{vmatrix} = \hat{i}(0)-\hat{j}(0)+\hat{k}(-2-0) = -2\hat{k}.
Area = |\vec{AB}\times\vec{AD}| = 2.
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