NCERT Solutions · Class 12 Mathematics Vector Algebra

Two nonzero vectors are collinear (parallel) precisely when one is a scalar multiple of the other, i.e. \(\displaystyle \vec{b}=\lambda\vec{a}\) for some scalar \(\displaystyle \lambda\); equivalently, their corresponding components are proportional.Let \(\displaystyle \vec{a}=2\hat{i}-3\hat{j}+4\hat{k}\) and \(\displaystyle \vec{b}=-4\hat{i}+6\hat{j}-8\hat{k}\). Compare the components: \[\frac{-4}{2}=-2,\qquad \frac{6}{-3}=-2,\qquad \frac{-8}{4}=-2\] All three ratios are the same number, so \[\vec{b}=-2\left(2\hat{i}-3\hat{j}+4\hat{k}\right)=-2\vec{a}\]Hence \(\displaystyle \vec{b}=\lambda\vec{a}\) with \(\displaystyle \lambda=-2\), so the two vectors are collinear. Since \(\displaystyle \lambda<0\) they lie along the same line but point in opposite directions, and \(\displaystyle |\vec{b}|=2|\vec{a}|\).

If \(\displaystyle \vec{r}=x\hat{i}+y\hat{j}+z\hat{k}\) makes angles \(\displaystyle \alpha,\beta,\gamma\) with the coordinate axes, its direction cosines are \[l=\cos\alpha=\frac{x}{|\vec{r}|},\quad m=\cos\beta=\frac{y}{|\vec{r}|},\quad n=\cos\gamma=\frac{z}{|\vec{r}|}\] that is, they are the components of the unit vector \(\displaystyle \hat{r}\) — so the vector must be divided by its magnitude, not used as it stands. \[|\vec{r}|=\sqrt{1^{2}+2^{2}+3^{2}}=\sqrt{14}\] \[l=\frac{1}{\sqrt{14}},\qquad m=\frac{2}{\sqrt{14}},\qquad n=\frac{3}{\sqrt{14}}\] Check: \(\displaystyle l^{2}+m^{2}+n^{2}=\frac{1+4+9}{14}=1\).Direction cosines: \(\displaystyle \left(\frac{1}{\sqrt{14}},\ \frac{2}{\sqrt{14}},\ \frac{3}{\sqrt{14}}\right)\).

"Directed from A to B" fixes the order: use \(\displaystyle \overrightarrow{\mathrm{AB}}=\) B \(\displaystyle -\) A. Taking B to A instead would reverse the sign of all three direction cosines. \[\overrightarrow{\mathrm{AB}}=(-1-1)\hat{i}+(-2-2)\hat{j}+\big(1-(-3)\big)\hat{k}=-2\hat{i}-4\hat{j}+4\hat{k}\] \[|\overrightarrow{\mathrm{AB}}|=\sqrt{(-2)^{2}+(-4)^{2}+4^{2}}=\sqrt{4+16+16}=\sqrt{36}=6\] The direction cosines are the components divided by this magnitude: \[l=\frac{-2}{6}=-\frac{1}{3},\qquad m=\frac{-4}{6}=-\frac{2}{3},\qquad n=\frac{4}{6}=\frac{2}{3}\] Check: \(\displaystyle \frac{1}{9}+\frac{4}{9}+\frac{4}{9}=1\).Direction cosines: \(\displaystyle \left(-\frac{1}{3},\ -\frac{2}{3},\ \frac{2}{3}\right)\).

Let \(\displaystyle \vec{r}=\hat{i}+\hat{j}+\hat{k}\) make angles \(\displaystyle \alpha,\beta,\gamma\) with OX, OY and OZ. By the definition of direction cosines, \(\displaystyle \cos\alpha=\dfrac{x}{|\vec{r}|}\), \(\displaystyle \cos\beta=\dfrac{y}{|\vec{r}|}\), \(\displaystyle \cos\gamma=\dfrac{z}{|\vec{r}|}\). \[|\vec{r}|=\sqrt{1^{2}+1^{2}+1^{2}}=\sqrt{3}\] \[\cos\alpha=\frac{1}{\sqrt{3}},\qquad \cos\beta=\frac{1}{\sqrt{3}},\qquad \cos\gamma=\frac{1}{\sqrt{3}}\] The three cosines are equal, and the angles a vector makes with the axes satisfy \(\displaystyle 0\le\alpha,\beta,\gamma\le\pi\), an interval on which cosine is one-to-one. Hence equal cosines force equal angles: \[\alpha=\beta=\gamma=\cos^{-1}\!\left(\frac{1}{\sqrt{3}}\right)\approx 54.74^{\circ}\]Therefore \(\displaystyle \hat{i}+\hat{j}+\hat{k}\) is equally inclined to the axes OX, OY and OZ.

The mid point is the section formula with ratio \(\displaystyle 1:1\): \[\vec{r}=\frac{\vec{p}+\vec{q}}{2}\] With P\(\displaystyle (2,3,4)\) and Q\(\displaystyle (4,1,-2)\), i.e. \(\displaystyle \vec{p}=2\hat{i}+3\hat{j}+4\hat{k}\) and \(\displaystyle \vec{q}=4\hat{i}+\hat{j}-2\hat{k}\), \[\vec{r}=\frac{(2+4)\hat{i}+(3+1)\hat{j}+(4-2)\hat{k}}{2}=\frac{6\hat{i}+4\hat{j}+2\hat{k}}{2}=3\hat{i}+2\hat{j}+\hat{k}\]Position vector of the mid point: \(\displaystyle 3\hat{i}+2\hat{j}+\hat{k}\), i.e. the point \(\displaystyle (3,2,1)\).

Form the side vectors as (terminal position vector \(\displaystyle -\) initial position vector), then apply the converse of Pythagoras' theorem to their squared lengths. \[\overrightarrow{\mathrm{AB}}=\vec{b}-\vec{a}=(2-3)\hat{i}+(-1+4)\hat{j}+(1+4)\hat{k}=-\hat{i}+3\hat{j}+5\hat{k}\] \[\overrightarrow{\mathrm{BC}}=\vec{c}-\vec{b}=(1-2)\hat{i}+(-3+1)\hat{j}+(-5-1)\hat{k}=-\hat{i}-2\hat{j}-6\hat{k}\] \[\overrightarrow{\mathrm{CA}}=\vec{a}-\vec{c}=(3-1)\hat{i}+(-4+3)\hat{j}+(-4+5)\hat{k}=2\hat{i}-\hat{j}+\hat{k}\] Squared lengths (keep them squared — no surds are needed): \[|\overrightarrow{\mathrm{AB}}|^{2}=1+9+25=35,\qquad |\overrightarrow{\mathrm{BC}}|^{2}=1+4+36=41,\qquad |\overrightarrow{\mathrm{CA}}|^{2}=4+1+1=6\] First check that a triangle exists: \(\displaystyle \overrightarrow{\mathrm{AB}}=-\hat{i}+3\hat{j}+5\hat{k}\) is not a scalar multiple of \(\displaystyle \overrightarrow{\mathrm{CA}}=2\hat{i}-\hat{j}+\hat{k}\) (since \(\displaystyle \tfrac{-1}{2}\neq\tfrac{3}{-1}\)), so A, B, C are not collinear.Now \[|\overrightarrow{\mathrm{AB}}|^{2}+|\overrightarrow{\mathrm{CA}}|^{2}=35+6=41=|\overrightarrow{\mathrm{BC}}|^{2}\] By the converse of Pythagoras' theorem, the angle opposite the longest side BC — the angle at vertex A — is a right angle.Hence A, B and C are the vertices of a right angled triangle, right angled at A.