NCERT Solutions · Class 12 Mathematics Vector Algebra

By the definition of the scalar (dot) product, \(\displaystyle \vec{a}\cdot\vec{b}=|\vec{a}|\,|\vec{b}|\cos\theta\), where \(\displaystyle \theta\) is the angle between the two vectors and \(\displaystyle 0\le\theta\le\pi\).Substituting \(\displaystyle |\vec{a}|=\sqrt{3}\), \(\displaystyle |\vec{b}|=2\) and \(\displaystyle \vec{a}\cdot\vec{b}=\sqrt{6}\), \[\sqrt{6}=\sqrt{3}\times 2\times\cos\theta\] \[\cos\theta=\frac{\sqrt{6}}{2\sqrt{3}}=\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}\] Since \(\displaystyle \theta\) is restricted to \(\displaystyle [0,\pi]\), the only angle whose cosine is \(\displaystyle \dfrac{1}{\sqrt{2}}\) is \(\displaystyle \dfrac{\pi}{4}\).The angle between \(\displaystyle \vec{a}\) and \(\displaystyle \vec{b}\) is \(\displaystyle \dfrac{\pi}{4}\) (that is, \(\displaystyle 45^{\circ}\)).

Let \(\displaystyle \vec{a}=\hat{i}-2\hat{j}+3\hat{k}\) and \(\displaystyle \vec{b}=3\hat{i}-2\hat{j}+\hat{k}\). Use \(\displaystyle \cos\theta=\dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}|\,|\vec{b}|}\), the dot product being computed componentwise.\[\vec{a}\cdot\vec{b}=(1)(3)+(-2)(-2)+(3)(1)=3+4+3=10\] \[|\vec{a}|=\sqrt{1^{2}+(-2)^{2}+3^{2}}=\sqrt{14},\qquad |\vec{b}|=\sqrt{3^{2}+(-2)^{2}+1^{2}}=\sqrt{14}\] Therefore \[\cos\theta=\frac{10}{\sqrt{14}\cdot\sqrt{14}}=\frac{10}{14}=\frac{5}{7}\] The required angle is \(\displaystyle \theta=\cos^{-1}\left(\dfrac{5}{7}\right)\).

The projection of \(\displaystyle \vec{a}\) on \(\displaystyle \vec{b}\) is \(\displaystyle \dfrac{\vec{a}\cdot\vec{b}}{|\vec{b}|}\). The step most often got wrong is which magnitude divides: it is the magnitude of the vector being projected onto, i.e. \(\displaystyle |\vec{b}|\).Here \(\displaystyle \vec{a}=\hat{i}-\hat{j}\) and \(\displaystyle \vec{b}=\hat{i}+\hat{j}\), so \[\vec{a}\cdot\vec{b}=(1)(1)+(-1)(1)=1-1=0,\qquad |\vec{b}|=\sqrt{1^{2}+1^{2}}=\sqrt{2}\] \[\text{Projection}=\frac{0}{\sqrt{2}}=0\] The projection is \(\displaystyle 0\). This is consistent: the dot product vanishing means the two vectors are perpendicular, so \(\displaystyle \hat{i}-\hat{j}\) has no component along \(\displaystyle \hat{i}+\hat{j}\).

Projection of \(\displaystyle \vec{a}\) on \(\displaystyle \vec{b}\) is \(\displaystyle \dfrac{\vec{a}\cdot\vec{b}}{|\vec{b}|}\), with \(\displaystyle \vec{a}=\hat{i}+3\hat{j}+7\hat{k}\) and \(\displaystyle \vec{b}=7\hat{i}-\hat{j}+8\hat{k}\).\[\vec{a}\cdot\vec{b}=(1)(7)+(3)(-1)+(7)(8)=7-3+56=60\] \[|\vec{b}|=\sqrt{7^{2}+(-1)^{2}+8^{2}}=\sqrt{49+1+64}=\sqrt{114}\] Hence \[\text{Projection}=\frac{60}{\sqrt{114}}=\frac{60\sqrt{114}}{114}=\frac{10\sqrt{114}}{19}\] The projection is \(\displaystyle \dfrac{60}{\sqrt{114}}\;\left(=\dfrac{10\sqrt{114}}{19}\right)\).

A vector is a unit vector when its magnitude equals \(\displaystyle 1\); two nonzero vectors are perpendicular when their dot product is \(\displaystyle 0\).Write \(\displaystyle \vec{a}=\dfrac{1}{7}(2\hat{i}+3\hat{j}+6\hat{k})\), \(\displaystyle \vec{b}=\dfrac{1}{7}(3\hat{i}-6\hat{j}+2\hat{k})\), \(\displaystyle \vec{c}=\dfrac{1}{7}(6\hat{i}+2\hat{j}-3\hat{k})\). A scalar multiple pulls out of the magnitude as \(\displaystyle |k\vec{v}|=|k||\vec{v}|\).\[|\vec{a}|=\frac{1}{7}\sqrt{2^{2}+3^{2}+6^{2}}=\frac{1}{7}\sqrt{4+9+36}=\frac{\sqrt{49}}{7}=\frac{7}{7}=1\] \[|\vec{b}|=\frac{1}{7}\sqrt{3^{2}+(-6)^{2}+2^{2}}=\frac{1}{7}\sqrt{9+36+4}=\frac{7}{7}=1\] \[|\vec{c}|=\frac{1}{7}\sqrt{6^{2}+2^{2}+(-3)^{2}}=\frac{1}{7}\sqrt{36+4+9}=\frac{7}{7}=1\] So each of the three is a unit vector.For the dot products, the two factors of \(\displaystyle \dfrac{1}{7}\) give a common \(\displaystyle \dfrac{1}{49}\): \[\vec{a}\cdot\vec{b}=\frac{1}{49}\big[(2)(3)+(3)(-6)+(6)(2)\big]=\frac{1}{49}(6-18+12)=0\] \[\vec{b}\cdot\vec{c}=\frac{1}{49}\big[(3)(6)+(-6)(2)+(2)(-3)\big]=\frac{1}{49}(18-12-6)=0\] \[\vec{c}\cdot\vec{a}=\frac{1}{49}\big[(6)(2)+(2)(3)+(-3)(6)\big]=\frac{1}{49}(12+6-18)=0\] Every pair has zero dot product and none of the vectors is \(\displaystyle \vec{0}\); hence the three given vectors are mutually perpendicular unit vectors.

Expand the product using distributivity of the dot product together with its commutativity \(\displaystyle \vec{a}\cdot\vec{b}=\vec{b}\cdot\vec{a}\) and \(\displaystyle \vec{v}\cdot\vec{v}=|\vec{v}|^{2}\): \[(\vec{a}+\vec{b})\cdot(\vec{a}-\vec{b})=\vec{a}\cdot\vec{a}-\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{a}-\vec{b}\cdot\vec{b}=|\vec{a}|^{2}-|\vec{b}|^{2}\] So the first condition is \[|\vec{a}|^{2}-|\vec{b}|^{2}=8\] Now substitute \(\displaystyle |\vec{a}|=8|\vec{b}|\): \[(8|\vec{b}|)^{2}-|\vec{b}|^{2}=8\ \Rightarrow\ 64|\vec{b}|^{2}-|\vec{b}|^{2}=8\ \Rightarrow\ 63|\vec{b}|^{2}=8\ \Rightarrow\ |\vec{b}|^{2}=\frac{8}{63}\] A magnitude is non-negative, so only the positive square root is admissible: \[|\vec{b}|=\sqrt{\frac{8}{63}}=\frac{2\sqrt{2}}{3\sqrt{7}}=\frac{2\sqrt{14}}{21}\] \[|\vec{a}|=8|\vec{b}|=\frac{16\sqrt{14}}{21}\] Hence \(\displaystyle |\vec{a}|=\dfrac{16\sqrt{14}}{21}\) and \(\displaystyle |\vec{b}|=\dfrac{2\sqrt{14}}{21}\).

Expand using distributivity of the dot product over addition, then use \(\displaystyle \vec{a}\cdot\vec{a}=|\vec{a}|^{2}\), \(\displaystyle \vec{b}\cdot\vec{b}=|\vec{b}|^{2}\) and commutativity \(\displaystyle \vec{b}\cdot\vec{a}=\vec{a}\cdot\vec{b}\): \[(3\vec{a}-5\vec{b})\cdot(2\vec{a}+7\vec{b})=3\vec{a}\cdot 2\vec{a}+3\vec{a}\cdot 7\vec{b}-5\vec{b}\cdot 2\vec{a}-5\vec{b}\cdot 7\vec{b}\] \[=6(\vec{a}\cdot\vec{a})+21(\vec{a}\cdot\vec{b})-10(\vec{b}\cdot\vec{a})-35(\vec{b}\cdot\vec{b})\] The middle two terms may be combined only because \(\displaystyle \vec{b}\cdot\vec{a}=\vec{a}\cdot\vec{b}\): \[=6|\vec{a}|^{2}+(21-10)\,\vec{a}\cdot\vec{b}-35|\vec{b}|^{2}\] \[=6|\vec{a}|^{2}+11\,\vec{a}\cdot\vec{b}-35|\vec{b}|^{2}\] The value of the product is \(\displaystyle 6|\vec{a}|^{2}+11\,\vec{a}\cdot\vec{b}-35|\vec{b}|^{2}\).

Let the common magnitude be \(\displaystyle |\vec{a}|=|\vec{b}|=k\), where \(\displaystyle k>0\). By the definition \(\displaystyle \vec{a}\cdot\vec{b}=|\vec{a}|\,|\vec{b}|\cos\theta\) with \(\displaystyle \theta=60^{\circ}\) and \(\displaystyle \vec{a}\cdot\vec{b}=\dfrac{1}{2}\), \[\frac{1}{2}=k\cdot k\cdot\cos 60^{\circ}=k^{2}\cdot\frac{1}{2}\] \[k^{2}=1\ \Rightarrow\ k=\pm 1\] A magnitude cannot be negative, so \(\displaystyle k=-1\) is rejected and \(\displaystyle k=1\).Hence \(\displaystyle |\vec{a}|=|\vec{b}|=1\).

Expand the product, using distributivity and \(\displaystyle \vec{x}\cdot\vec{a}=\vec{a}\cdot\vec{x}\): \[(\vec{x}-\vec{a})\cdot(\vec{x}+\vec{a})=\vec{x}\cdot\vec{x}+\vec{x}\cdot\vec{a}-\vec{a}\cdot\vec{x}-\vec{a}\cdot\vec{a}=|\vec{x}|^{2}-|\vec{a}|^{2}\] Since \(\displaystyle \vec{a}\) is a unit vector, \(\displaystyle |\vec{a}|=1\), so \(\displaystyle |\vec{a}|^{2}=1\). Therefore \[|\vec{x}|^{2}-1=12\ \Rightarrow\ |\vec{x}|^{2}=13\] Taking the positive root (a magnitude is non-negative), \[|\vec{x}|=\sqrt{13}\]

Two vectors are perpendicular exactly when their dot product is zero. First form \(\displaystyle \vec{a}+\lambda\vec{b}\) with \(\displaystyle \vec{a}=2\hat{i}+2\hat{j}+3\hat{k}\) and \(\displaystyle \vec{b}=-\hat{i}+2\hat{j}+\hat{k}\): \[\vec{a}+\lambda\vec{b}=(2-\lambda)\hat{i}+(2+2\lambda)\hat{j}+(3+\lambda)\hat{k}\] Note that \(\displaystyle \vec{c}=3\hat{i}+\hat{j}\) has \(\displaystyle \hat{k}\)-component \(\displaystyle 0\), so that term drops out of the dot product. Setting \(\displaystyle (\vec{a}+\lambda\vec{b})\cdot\vec{c}=0\): \[3(2-\lambda)+1(2+2\lambda)+0(3+\lambda)=0\] \[6-3\lambda+2+2\lambda=0\] \[8-\lambda=0\] \[\lambda=8\]