NCERT Solutions · Class 12 Mathematics Vector Algebra

A unit vector lying in the XY-plane has no \(\displaystyle \hat k\) component, and if it makes an angle \(\displaystyle \theta\) with the positive \(\displaystyle x\)-axis its direction cosines along \(\displaystyle x\) and \(\displaystyle y\) are \(\displaystyle \cos\theta\) and \(\displaystyle \sin\theta\). So \[\hat r=\cos\theta\,\hat i+\sin\theta\,\hat j.\]Putting \(\displaystyle \theta=30^{\circ}\), \[\hat r=\cos 30^{\circ}\,\hat i+\sin 30^{\circ}\,\hat j=\frac{\sqrt{3}}{2}\hat i+\frac{1}{2}\hat j.\]Check: \(\displaystyle \left(\frac{\sqrt{3}}{2}\right)^{2}+\left(\frac{1}{2}\right)^{2}=\frac{3}{4}+\frac{1}{4}=1\), so \(\displaystyle |\hat r|=1\).Required unit vector: \(\displaystyle \dfrac{\sqrt{3}}{2}\hat i+\dfrac{1}{2}\hat j\).

For any two points, the vector joining them is (position vector of the terminal point) \(\displaystyle -\) (position vector of the initial point): \[\overrightarrow{\mathrm{PQ}}=\overrightarrow{\mathrm{OQ}}-\overrightarrow{\mathrm{OP}}.\]Here \(\displaystyle \overrightarrow{\mathrm{OP}}=x_{1}\hat i+y_{1}\hat j+z_{1}\hat k\) and \(\displaystyle \overrightarrow{\mathrm{OQ}}=x_{2}\hat i+y_{2}\hat j+z_{2}\hat k\), so \[\overrightarrow{\mathrm{PQ}}=\left(x_{2}-x_{1}\right)\hat i+\left(y_{2}-y_{1}\right)\hat j+\left(z_{2}-z_{1}\right)\hat k.\]The scalar components are the coefficients of \(\displaystyle \hat i,\hat j,\hat k\): \[x_{2}-x_{1},\qquad y_{2}-y_{1},\qquad z_{2}-z_{1}.\]The magnitude is the square root of the sum of their squares: \[\left|\overrightarrow{\mathrm{PQ}}\right|=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}}.\]

Take \(\displaystyle \hat i\) pointing due east and \(\displaystyle \hat j\) due north, with O her starting point.First leg, $\displaystyle 4$ km west (the direction \(\displaystyle -\hat i\)): \[\overrightarrow{\mathrm{OA}}=-4\hat i.\]Second leg, $\displaystyle 3$ km at \(\displaystyle 30^{\circ}\) east of north. "East of north" means the angle is measured from the north direction turning towards the east, so this direction makes \(\displaystyle 30^{\circ}\) with \(\displaystyle \hat j\) — the north component carries the cosine and the east component the sine: \[\overrightarrow{\mathrm{AB}}=3\sin 30^{\circ}\,\hat i+3\cos 30^{\circ}\,\hat j=\frac{3}{2}\hat i+\frac{3\sqrt{3}}{2}\hat j.\]By the triangle law of addition, the displacement from the start is \[\overrightarrow{\mathrm{OB}}=\overrightarrow{\mathrm{OA}}+\overrightarrow{\mathrm{AB}}=\left(-4+\frac{3}{2}\right)\hat i+\frac{3\sqrt{3}}{2}\hat j=-\frac{5}{2}\hat i+\frac{3\sqrt{3}}{2}\hat j.\]Its magnitude: \[\left|\overrightarrow{\mathrm{OB}}\right|=\sqrt{\left(-\frac{5}{2}\right)^{2}+\left(\frac{3\sqrt{3}}{2}\right)^{2}}=\sqrt{\frac{25}{4}+\frac{27}{4}}=\sqrt{\frac{52}{4}}=\sqrt{13}.\]Displacement \(\displaystyle =-\dfrac{5}{2}\hat i+\dfrac{3\sqrt{3}}{2}\hat j\), i.e. \(\displaystyle \sqrt{13}\approx 3.6\) km in a direction north-west of O (the \(\displaystyle \hat i\) component is negative, the \(\displaystyle \hat j\) component positive).

No, it is not true in general.With \(\displaystyle \theta\) the angle between \(\displaystyle \vec b\) and \(\displaystyle \vec c\), \[|\vec a|^{2}=|\vec b+\vec c|^{2}=(\vec b+\vec c)\cdot(\vec b+\vec c)=|\vec b|^{2}+|\vec c|^{2}+2|\vec b||\vec c|\cos\theta,\] whereas \[\left(|\vec b|+|\vec c|\right)^{2}=|\vec b|^{2}+|\vec c|^{2}+2|\vec b||\vec c|.\]Since \(\displaystyle \cos\theta\le 1\), we always have \(\displaystyle |\vec a|\le|\vec b|+|\vec c|\) (the triangle inequality), and the two sides are equal only when \(\displaystyle \cos\theta=1\), i.e. only when \(\displaystyle \vec b\) and \(\displaystyle \vec c\) are like vectors (collinear and in the same sense), or one of them is \(\displaystyle \vec 0\).Counterexample: take \(\displaystyle \vec b=\hat i\) and \(\displaystyle \vec c=-\hat i\). Then \(\displaystyle \vec a=\vec b+\vec c=\vec 0\), so \(\displaystyle |\vec a|=0\) but \(\displaystyle |\vec b|+|\vec c|=2\).Hence \(\displaystyle |\vec a|=|\vec b|+|\vec c|\) is not true in general; it holds exactly when \(\displaystyle \vec b\) and \(\displaystyle \vec c\) have the same direction, and otherwise \(\displaystyle |\vec a|<|\vec b|+|\vec c|\).

\[x(\hat i+\hat j+\hat k)=x\hat i+x\hat j+x\hat k,\] so by the formula for the magnitude of a vector, \[\left|x(\hat i+\hat j+\hat k)\right|=\sqrt{x^{2}+x^{2}+x^{2}}=\sqrt{3x^{2}}=\sqrt{3}\,|x|.\] Note \(\displaystyle \sqrt{x^{2}}=|x|\), not \(\displaystyle x\) — that is what keeps the negative root in play.For the vector to be a unit vector, \[\sqrt{3}\,|x|=1\ \Longrightarrow\ |x|=\frac{1}{\sqrt{3}}\ \Longrightarrow\ x=\pm\frac{1}{\sqrt{3}}.\]Hence \(\displaystyle x=\pm\dfrac{1}{\sqrt{3}}\).

The resultant of the two vectors is their sum: \[\vec a+\vec b=(2+1)\hat i+(3-2)\hat j+(-1+1)\hat k=3\hat i+\hat j.\] \[|\vec a+\vec b|=\sqrt{3^{2}+1^{2}+0^{2}}=\sqrt{10}.\]A unit vector along the resultant is \[\hat c=\frac{\vec a+\vec b}{|\vec a+\vec b|}=\frac{1}{\sqrt{10}}\left(3\hat i+\hat j\right).\]A vector of magnitude $\displaystyle 5$ parallel to it is \(\displaystyle \pm 5\hat c\) (both signs qualify as "parallel", one along the resultant and one opposite to it): \[\pm\frac{5}{\sqrt{10}}\left(3\hat i+\hat j\right)=\pm\left(\frac{15}{\sqrt{10}}\hat i+\frac{5}{\sqrt{10}}\hat j\right)=\pm\left(\frac{3\sqrt{10}}{2}\hat i+\frac{\sqrt{10}}{2}\hat j\right).\]Required vector: \(\displaystyle \pm\left(\dfrac{3\sqrt{10}}{2}\hat i+\dfrac{\sqrt{10}}{2}\hat j\right)\).

First form the combination term by term: \[2\vec a=2\hat i+2\hat j+2\hat k,\qquad -\vec b=-2\hat i+\hat j-3\hat k,\qquad 3\vec c=3\hat i-6\hat j+3\hat k.\] (The sign trap is \(\displaystyle -\vec b\): every component of \(\displaystyle \vec b\) flips, so \(\displaystyle -(-\hat j)=+\hat j\).)Adding, \[2\vec a-\vec b+3\vec c=(2-2+3)\hat i+(2+1-6)\hat j+(2-3+3)\hat k=3\hat i-3\hat j+2\hat k.\]Its magnitude: \[\left|2\vec a-\vec b+3\vec c\right|=\sqrt{3^{2}+(-3)^{2}+2^{2}}=\sqrt{9+9+4}=\sqrt{22}.\]A unit vector parallel to it is the vector divided by its magnitude: \[\pm\frac{1}{\sqrt{22}}\left(3\hat i-3\hat j+2\hat k\right)=\pm\left(\frac{3}{\sqrt{22}}\hat i-\frac{3}{\sqrt{22}}\hat j+\frac{2}{\sqrt{22}}\hat k\right).\]The required unit vector is \(\displaystyle \dfrac{3}{\sqrt{22}}\hat i-\dfrac{3}{\sqrt{22}}\hat j+\dfrac{2}{\sqrt{22}}\hat k\) (its negative is also parallel).

Three points are collinear if two of the vectors joining them are parallel, i.e. one is a scalar multiple of the other, since they share a common point.\[\overrightarrow{\mathrm{AB}}=(5-1)\hat i+(0-(-2))\hat j+(-2-(-8))\hat k=4\hat i+2\hat j+6\hat k,\] \[\overrightarrow{\mathrm{BC}}=(11-5)\hat i+(3-0)\hat j+(7-(-2))\hat k=6\hat i+3\hat j+9\hat k,\] \[\overrightarrow{\mathrm{AC}}=(11-1)\hat i+(3-(-2))\hat j+(7-(-8))\hat k=10\hat i+5\hat j+15\hat k.\]Observe \[\overrightarrow{\mathrm{AB}}=\frac{2}{3}\left(6\hat i+3\hat j+9\hat k\right)=\frac{2}{3}\overrightarrow{\mathrm{BC}},\] so \(\displaystyle \overrightarrow{\mathrm{AB}}\parallel\overrightarrow{\mathrm{BC}}\) and they have the common point B. Hence A, B, C are collinear.(Equivalently, \(\displaystyle |\overrightarrow{\mathrm{AB}}|=\sqrt{56}=2\sqrt{14}\), \(\displaystyle |\overrightarrow{\mathrm{BC}}|=\sqrt{126}=3\sqrt{14}\), \(\displaystyle |\overrightarrow{\mathrm{AC}}|=\sqrt{350}=5\sqrt{14}\), and \(\displaystyle |\overrightarrow{\mathrm{AB}}|+|\overrightarrow{\mathrm{BC}}|=|\overrightarrow{\mathrm{AC}}|\).)Since the scalar \(\displaystyle \frac{2}{3}\) is positive, \(\displaystyle \overrightarrow{\mathrm{AB}}\) and \(\displaystyle \overrightarrow{\mathrm{BC}}\) point the same way, so B lies between A and C and divides AC internally in the ratio \[\mathrm{AB}:\mathrm{BC}=2\sqrt{14}:3\sqrt{14}=2:3.\]B divides AC internally in the ratio \(\displaystyle 2:3\).

Section formula (external division): if R divides the line joining P and Q externally in the ratio \(\displaystyle m:n\), then \[\vec r=\frac{m\vec q-n\vec p}{m-n}.\]Here \(\displaystyle \vec p=2\vec a+\vec b\), \(\displaystyle \vec q=\vec a-3\vec b\), \(\displaystyle m=1\), \(\displaystyle n=2\). The denominator \(\displaystyle m-n=-1\) is negative — that sign is where the working usually goes wrong: \[\vec r=\frac{1\cdot\left(\vec a-3\vec b\right)-2\left(2\vec a+\vec b\right)}{1-2}=\frac{\vec a-3\vec b-4\vec a-2\vec b}{-1}=\frac{-3\vec a-5\vec b}{-1}=3\vec a+5\vec b.\]So the position vector of R is \(\displaystyle 3\vec a+5\vec b\).Now the midpoint of RQ has position vector \[\frac{\vec r+\vec q}{2}=\frac{\left(3\vec a+5\vec b\right)+\left(\vec a-3\vec b\right)}{2}=\frac{4\vec a+2\vec b}{2}=2\vec a+\vec b=\vec p.\]This is exactly the position vector of P, so P is the mid point of the line segment RQ.

Let the adjacent sides be \(\displaystyle \vec a=2\hat i-4\hat j+5\hat k\) and \(\displaystyle \vec b=\hat i-2\hat j-3\hat k\), drawn from the same vertex. Then the diagonals of the parallelogram are \(\displaystyle \vec a+\vec b\) and \(\displaystyle \vec a-\vec b\).Diagonal through the common vertex: \[\vec a+\vec b=3\hat i-6\hat j+2\hat k,\qquad |\vec a+\vec b|=\sqrt{9+36+4}=\sqrt{49}=7,\] so a unit vector parallel to it is \[\frac{1}{7}\left(3\hat i-6\hat j+2\hat k\right).\]The other diagonal: \[\vec a-\vec b=\hat i-2\hat j+8\hat k,\qquad |\vec a-\vec b|=\sqrt{1+4+64}=\sqrt{69},\] giving the unit vector \(\displaystyle \frac{1}{\sqrt{69}}\left(\hat i-2\hat j+8\hat k\right)\).Area of a parallelogram with adjacent sides \(\displaystyle \vec a,\vec b\) is \(\displaystyle |\vec a\times\vec b|\): \[\vec a\times\vec b=\left|\begin{array}{rrr}\hat i & \hat j & \hat k\\ 2 & -4 & 5\\ 1 & -2 & -3\end{array}\right|=\hat i\left(12+10\right)-\hat j\left(-6-5\right)+\hat k\left(-4+4\right)=22\hat i+11\hat j.\] \[|\vec a\times\vec b|=\sqrt{22^{2}+11^{2}}=\sqrt{484+121}=\sqrt{605}=11\sqrt{5}.\]Unit vector parallel to the diagonal \(\displaystyle \vec a+\vec b\): \(\displaystyle \dfrac{1}{7}\left(3\hat i-6\hat j+2\hat k\right)\) (the other diagonal gives \(\displaystyle \frac{1}{\sqrt{69}}\left(\hat i-2\hat j+8\hat k\right)\)); area \(\displaystyle =11\sqrt{5}\) square units.