NCERT Solutions · Class 12 Mathematics Vector Algebra

By definition, the cross product of \(\displaystyle \vec{a}=a_1\hat i+a_2\hat j+a_3\hat k\) and \(\displaystyle \vec{b}=b_1\hat i+b_2\hat j+b_3\hat k\) is the determinant \[\vec{a}\times\vec{b}=\left|\begin{array}{rrr}\hat i&\hat j&\hat k\\ a_1&a_2&a_3\\ b_1&b_2&b_3\end{array}\right|. \] Here \(\displaystyle \vec{a}=\hat i-7\hat j+7\hat k\) and \(\displaystyle \vec{b}=3\hat i-2\hat j+2\hat k\), so \[\vec{a}\times\vec{b}=\left|\begin{array}{rrr}\hat i&\hat j&\hat k\\ 1&-7&7\\ 3&-2&2\end{array}\right|. \] Expanding along the first row (note the minus sign attached to the \(\displaystyle \hat j\) term): \[\vec{a}\times\vec{b}=\hat i\big[(-7)(2)-(7)(-2)\big]-\hat j\big[(1)(2)-(7)(3)\big]+\hat k\big[(1)(-2)-(-7)(3)\big] \] \[=\hat i(-14+14)-\hat j(2-21)+\hat k(-2+21)=0\,\hat i+19\,\hat j+19\,\hat k. \] Hence \[|\vec{a}\times\vec{b}|=\sqrt{0^{2}+19^{2}+19^{2}}=\sqrt{2\times 361}=19\sqrt{2}. \] Final answer: \(\displaystyle |\vec{a}\times\vec{b}|=19\sqrt{2}\).

The cross product of two vectors is perpendicular to both of them, so \(\displaystyle (\vec{a}+\vec{b})\times(\vec{a}-\vec{b})\) is perpendicular to \(\displaystyle \vec{a}+\vec{b}\) and to \(\displaystyle \vec{a}-\vec{b}\); dividing it by its own magnitude makes it a unit vector.With \(\displaystyle \vec{a}=3\hat i+2\hat j+2\hat k\) and \(\displaystyle \vec{b}=\hat i+2\hat j-2\hat k\), \[\vec{a}+\vec{b}=4\hat i+4\hat j+0\hat k,\qquad \vec{a}-\vec{b}=2\hat i+0\hat j+4\hat k. \] Then \[(\vec{a}+\vec{b})\times(\vec{a}-\vec{b})=\left|\begin{array}{rrr}\hat i&\hat j&\hat k\\ 4&4&0\\ 2&0&4\end{array}\right| =\hat i(4\cdot 4-0\cdot 0)-\hat j(4\cdot 4-0\cdot 2)+\hat k(4\cdot 0-4\cdot 2) \] \[=16\,\hat i-16\,\hat j-8\,\hat k. \] Its magnitude is \[\sqrt{16^{2}+(-16)^{2}+(-8)^{2}}=\sqrt{256+256+64}=\sqrt{576}=24. \] Therefore the required unit vector is \[\frac{16\hat i-16\hat j-8\hat k}{24}=\frac{2}{3}\hat i-\frac{2}{3}\hat j-\frac{1}{3}\hat k. \] Final answer: \(\displaystyle \pm\dfrac{1}{3}\left(2\hat i-2\hat j-\hat k\right)\) — the negative of this vector is equally perpendicular to both, so either sign is acceptable.

If a unit vector makes angles \(\displaystyle \alpha,\beta,\gamma\) with \(\displaystyle \hat i,\hat j,\hat k\), then \(\displaystyle \cos\alpha,\cos\beta,\cos\gamma\) are its direction cosines and they satisfy \[\cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma=1 . \] Here \(\displaystyle \alpha=\dfrac{\pi}{3}\), \(\displaystyle \beta=\dfrac{\pi}{4}\), \(\displaystyle \gamma=\theta\), so \[\cos^{2}\frac{\pi}{3}+\cos^{2}\frac{\pi}{4}+\cos^{2}\theta=1 \quad\Longrightarrow\quad \left(\frac{1}{2}\right)^{2}+\left(\frac{1}{\sqrt{2}}\right)^{2}+\cos^{2}\theta=1 . \] That is \[\frac{1}{4}+\frac{1}{2}+\cos^{2}\theta=1\quad\Longrightarrow\quad \cos^{2}\theta=\frac{1}{4}\quad\Longrightarrow\quad\cos\theta=\pm\frac{1}{2}. \] The sign is decided by the condition that \(\displaystyle \theta\) is acute: an acute angle has a positive cosine, so \(\displaystyle \cos\theta=\dfrac12\) and \[\theta=\frac{\pi}{3}. \] The components of a unit vector are exactly its direction cosines, so \[\vec{a}=\cos\frac{\pi}{3}\,\hat i+\cos\frac{\pi}{4}\,\hat j+\cos\frac{\pi}{3}\,\hat k=\frac{1}{2}\hat i+\frac{1}{\sqrt{2}}\hat j+\frac{1}{2}\hat k . \] Final answer: \(\displaystyle \theta=\dfrac{\pi}{3}\) and \(\displaystyle \vec{a}=\dfrac12\hat i+\dfrac{1}{\sqrt2}\hat j+\dfrac12\hat k\), i.e. the components are \(\displaystyle \left\langle \dfrac12,\dfrac{1}{\sqrt2},\dfrac12\right\rangle\).

Use the distributive property of the cross product over addition, together with the two facts \(\displaystyle \vec{u}\times\vec{u}=\vec{0}\) (the cross product of a vector with itself vanishes) and \(\displaystyle \vec{v}\times\vec{u}=-(\vec{u}\times\vec{v})\) (the cross product is anti-commutative).Expanding the left side, being careful to keep the order of the factors in each term: \[(\vec{a}-\vec{b})\times(\vec{a}+\vec{b})=\vec{a}\times\vec{a}+\vec{a}\times\vec{b}-\vec{b}\times\vec{a}-\vec{b}\times\vec{b}. \] Now \(\displaystyle \vec{a}\times\vec{a}=\vec{0}\) and \(\displaystyle \vec{b}\times\vec{b}=\vec{0}\), and by anti-commutativity \(\displaystyle -\vec{b}\times\vec{a}=+\,\vec{a}\times\vec{b}\). Hence \[(\vec{a}-\vec{b})\times(\vec{a}+\vec{b})=\vec{0}+\vec{a}\times\vec{b}+\vec{a}\times\vec{b}-\vec{0}=2(\vec{a}\times\vec{b}). \] Hence \(\displaystyle (\vec{a}-\vec{b})\times(\vec{a}+\vec{b})=2(\vec{a}\times\vec{b})\), as required.(Geometrically: the diagonals of the parallelogram on \(\displaystyle \vec a,\vec b\) span a parallelogram of twice its area.)

Compute the cross product as a determinant and set every component equal to zero. \[(2\hat i+6\hat j+27\hat k)\times(\hat i+\lambda\hat j+\mu\hat k)=\left|\begin{array}{rrr}\hat i&\hat j&\hat k\\ 2&6&27\\ 1&\lambda&\mu\end{array}\right| \] \[=\hat i\,(6\mu-27\lambda)-\hat j\,(2\mu-27)+\hat k\,(2\lambda-6). \] A vector is \(\displaystyle \vec 0\) only when each of its components is \(\displaystyle 0\), so \[6\mu-27\lambda=0,\qquad 2\mu-27=0,\qquad 2\lambda-6=0 . \] The last two give directly \[\mu=\frac{27}{2},\qquad \lambda=3 . \] Check these in the first equation (this is the step worth not skipping — the three equations must be consistent): \[6\left(\frac{27}{2}\right)-27(3)=81-81=0.\;\checkmark \] Final answer: \(\displaystyle \lambda=3,\ \mu=\dfrac{27}{2}\).(Equivalently, a vanishing cross product of two non-zero vectors means they are parallel, so \(\displaystyle \dfrac{2}{1}=\dfrac{6}{\lambda}=\dfrac{27}{\mu}\), giving the same values.)

Suppose, if possible, that both \(\displaystyle \vec a\neq\vec 0\) and \(\displaystyle \vec b\neq\vec 0\), and let \(\displaystyle \theta\in[0,\pi]\) be the angle between them.From \(\displaystyle \vec a\cdot\vec b=|\vec a||\vec b|\cos\theta=0\) with \(\displaystyle |\vec a|\neq 0,\ |\vec b|\neq 0\), we get \[\cos\theta=0\quad\Longrightarrow\quad \theta=\frac{\pi}{2}. \] From \(\displaystyle |\vec a\times\vec b|=|\vec a||\vec b|\sin\theta=0\) with the same non-zero magnitudes, \[\sin\theta=0\quad\Longrightarrow\quad \theta=0\ \text{or}\ \theta=\pi . \] These two conclusions contradict each other: \(\displaystyle \theta\) cannot be \(\displaystyle \pi/2\) and simultaneously \(\displaystyle 0\) or \(\displaystyle \pi\). So the supposition is false.Conclusion: at least one of \(\displaystyle \vec a\) and \(\displaystyle \vec b\) must be the zero vector, i.e. \(\displaystyle \vec a=\vec 0\) or \(\displaystyle \vec b=\vec 0\) (possibly both).

Write \(\displaystyle \vec b+\vec c=(b_1+c_1)\hat i+(b_2+c_2)\hat j+(b_3+c_3)\hat k\) and expand the cross product as a determinant. \[\vec a\times(\vec b+\vec c)=\left|\begin{array}{ccc}\hat i&\hat j&\hat k\\ a_1&a_2&a_3\\ b_1+c_1&b_2+c_2&b_3+c_3\end{array}\right| \] \[=\hat i\big[a_2(b_3+c_3)-a_3(b_2+c_2)\big]-\hat j\big[a_1(b_3+c_3)-a_3(b_1+c_1)\big]+\hat k\big[a_1(b_2+c_2)-a_2(b_1+c_1)\big]. \] Now split each bracket into a \(\displaystyle b\)-part and a \(\displaystyle c\)-part and regroup: \[=\Big\{\hat i(a_2b_3-a_3b_2)-\hat j(a_1b_3-a_3b_1)+\hat k(a_1b_2-a_2b_1)\Big\} +\Big\{\hat i(a_2c_3-a_3c_2)-\hat j(a_1c_3-a_3c_1)+\hat k(a_1c_2-a_2c_1)\Big\}. \] The first brace is precisely \[\left|\begin{array}{ccc}\hat i&\hat j&\hat k\\ a_1&a_2&a_3\\ b_1&b_2&b_3\end{array}\right|=\vec a\times\vec b, \] and the second brace is precisely \[\left|\begin{array}{ccc}\hat i&\hat j&\hat k\\ a_1&a_2&a_3\\ c_1&c_2&c_3\end{array}\right|=\vec a\times\vec c . \] Hence \[\vec a\times(\vec b+\vec c)=\vec a\times\vec b+\vec a\times\vec c, \] which is the required distributive law.

The converse would read: "if \(\displaystyle \vec a\times\vec b=\vec 0\), then \(\displaystyle \vec a=\vec 0\) or \(\displaystyle \vec b=\vec 0\)." This is not true.Reason: \(\displaystyle |\vec a\times\vec b|=|\vec a||\vec b|\sin\theta\), and this vanishes not only when a magnitude is zero but also when \(\displaystyle \sin\theta=0\), i.e. when the two non-zero vectors are parallel or anti-parallel.Counter-example. Take \[\vec a=2\hat i+3\hat j+4\hat k,\qquad \vec b=4\hat i+6\hat j+8\hat k\ (=2\vec a). \] Both are clearly non-zero, yet \[\vec a\times\vec b=\left|\begin{array}{rrr}\hat i&\hat j&\hat k\\ 2&3&4\\ 4&6&8\end{array}\right| =\hat i(3\cdot 8-4\cdot 6)-\hat j(2\cdot 8-4\cdot 4)+\hat k(2\cdot 6-3\cdot 4) \] \[=\hat i(24-24)-\hat j(16-16)+\hat k(12-12)=\vec 0 . \] So \(\displaystyle \vec a\times\vec b=\vec 0\) with neither vector zero.Conclusion: the converse is false. The correct statement is that \(\displaystyle \vec a\times\vec b=\vec 0\) if and only if \(\displaystyle \vec a=\vec 0\), or \(\displaystyle \vec b=\vec 0\), or \(\displaystyle \vec a\) and \(\displaystyle \vec b\) are parallel.

If two sides of a triangle are the vectors \(\displaystyle \overrightarrow{AB}\) and \(\displaystyle \overrightarrow{AC}\), its area is half the area of the parallelogram on them: \[\text{Area}=\frac{1}{2}\left|\overrightarrow{AB}\times\overrightarrow{AC}\right| . \] With \(\displaystyle A(1,1,2)\), \(\displaystyle B(2,3,5)\), \(\displaystyle C(1,5,5)\), \[\overrightarrow{AB}=(2-1)\hat i+(3-1)\hat j+(5-2)\hat k=\hat i+2\hat j+3\hat k, \] \[\overrightarrow{AC}=(1-1)\hat i+(5-1)\hat j+(5-2)\hat k=0\,\hat i+4\hat j+3\hat k . \] Then \[\overrightarrow{AB}\times\overrightarrow{AC}=\left|\begin{array}{rrr}\hat i&\hat j&\hat k\\ 1&2&3\\ 0&4&3\end{array}\right| =\hat i(2\cdot 3-3\cdot 4)-\hat j(1\cdot 3-3\cdot 0)+\hat k(1\cdot 4-2\cdot 0) \] \[=-6\,\hat i-3\,\hat j+4\,\hat k . \] Its magnitude is \[\left|\overrightarrow{AB}\times\overrightarrow{AC}\right|=\sqrt{(-6)^{2}+(-3)^{2}+4^{2}}=\sqrt{36+9+16}=\sqrt{61}. \] Final answer: area of \(\displaystyle \triangle ABC=\dfrac{\sqrt{61}}{2}\) square units.

For a parallelogram whose adjacent sides are the vectors \(\displaystyle \vec a\) and \(\displaystyle \vec b\), \[\text{Area}=|\vec a\times\vec b| . \] With \(\displaystyle \vec a=\hat i-\hat j+3\hat k\) and \(\displaystyle \vec b=2\hat i-7\hat j+\hat k\), \[\vec a\times\vec b=\left|\begin{array}{rrr}\hat i&\hat j&\hat k\\ 1&-1&3\\ 2&-7&1\end{array}\right| =\hat i\big[(-1)(1)-(3)(-7)\big]-\hat j\big[(1)(1)-(3)(2)\big]+\hat k\big[(1)(-7)-(-1)(2)\big] \] \[=\hat i(-1+21)-\hat j(1-6)+\hat k(-7+2)=20\,\hat i+5\,\hat j-5\,\hat k . \] Hence \[|\vec a\times\vec b|=\sqrt{20^{2}+5^{2}+(-5)^{2}}=\sqrt{400+25+25}=\sqrt{450}=15\sqrt{2}. \] Final answer: the area of the parallelogram is \(\displaystyle 15\sqrt{2}\) square units.

Use \(\displaystyle |\vec a\times\vec b|=|\vec a||\vec b|\sin\theta\), where \(\displaystyle \theta\in[0,\pi]\) is the angle between the vectors."\(\displaystyle \vec a\times\vec b\) is a unit vector" means \(\displaystyle |\vec a\times\vec b|=1\). With \(\displaystyle |\vec a|=3\) and \(\displaystyle |\vec b|=\dfrac{\sqrt2}{3}\), \[3\cdot\frac{\sqrt2}{3}\cdot\sin\theta=1\quad\Longrightarrow\quad \sqrt{2}\,\sin\theta=1\quad\Longrightarrow\quad \sin\theta=\frac{1}{\sqrt2}. \] For \(\displaystyle 0\le\theta\le\pi\) this gives \(\displaystyle \theta=\dfrac{\pi}{4}\) or \(\displaystyle \theta=\dfrac{3\pi}{4}\); only \(\displaystyle \dfrac{\pi}{4}\) appears among the options.Final answer: option (B) \(\displaystyle \dfrac{\pi}{4}\).

Take the position vectors as points: \[A\left(-1,\tfrac12,4\right),\quad B\left(1,\tfrac12,4\right),\quad C\left(1,-\tfrac12,4\right),\quad D\left(-1,-\tfrac12,4\right). \] The adjacent sides are \[\overrightarrow{AB}=\big(1-(-1)\big)\hat i+\left(\tfrac12-\tfrac12\right)\hat j+(4-4)\hat k=2\hat i, \] \[\overrightarrow{BC}=(1-1)\hat i+\left(-\tfrac12-\tfrac12\right)\hat j+(4-4)\hat k=-\hat j . \] (Indeed \(\displaystyle \overrightarrow{AB}\cdot\overrightarrow{BC}=0\), confirming the right angle at \(\displaystyle B\), so the figure really is a rectangle.)The area of the rectangle with adjacent sides \(\displaystyle \overrightarrow{AB}\) and \(\displaystyle \overrightarrow{BC}\) is \[\left|\overrightarrow{AB}\times\overrightarrow{BC}\right|=\left|\ \left|\begin{array}{rrr}\hat i&\hat j&\hat k\\ 2&0&0\\ 0&-1&0\end{array}\right|\ \right| =\left|\hat i(0-0)-\hat j(0-0)+\hat k(-2-0)\right|=|-2\hat k|=2 . \] Final answer: option (C) \(\displaystyle 2\).