Class 12 Maths NCERT Solutions Chapter 10 Ex 10.2 – Addition, Components and Section Formula

For \vec{r}=x\hat{i}+y\hat{j}+z\hat{k}, the magnitude is |\vec{r}|=\sqrt{x^2+y^2+z^2}.

|\vec{a}| = \sqrt{1^2+1^2+1^2} = \sqrt3.

|\vec{b}| = \sqrt{2^2+(-7)^2+(-3)^2} = \sqrt{4+49+9} = \sqrt{62}.

|\vec{c}| = \sqrt{\left(\frac{1}{\sqrt3}\right)^2+\left(\frac{1}{\sqrt3}\right)^2+\left(-\frac{1}{\sqrt3}\right)^2} = \sqrt{\frac13+\frac13+\frac13} = 1.

Consider \vec{a}=\hat{i}+2\hat{j}+2\hat{k} and \vec{b}=2\hat{i}+2\hat{j}+\hat{k}.

|\vec{a}|=\sqrt{1+4+4}=3 and |\vec{b}|=\sqrt{4+4+1}=3.

Consider \vec{a}=\hat{i}+\hat{j}+\hat{k} and \vec{b}=2\hat{i}+2\hat{j}+2\hat{k}.

Here \vec{b}=2\vec{a}, so the direction ratios of both vectors are in the same proportion (1 : 1 : 1), meaning they point in exactly the same direction, though |\vec{a}|=\sqrt3 while |\vec{b}|=2\sqrt3.

Two vectors are equal if and only if their corresponding components are equal, so matching the coefficients of \hat{i} and \hat{j} between 2\hat{i}+3\hat{j} and x\hat{i}+y\hat{j} gives x=2 and y=3 directly.

For initial point P(2, 1) and terminal point Q(−5, 7): \vec{PQ} = (-5-2)\hat{i} + (7-1)\hat{j} = -7\hat{i}+6\hat{j}.

Adding the respective components: \vec{a}+\vec{b}+\vec{c} = (1-2+1)\hat{i} + (-2+4-6)\hat{j} + (1+5-7)\hat{k}.

|\vec{a}| = \sqrt{1^2+1^2+2^2} = \sqrt6.

\hat{a} = \dfrac{1}{|\vec{a}|}\vec{a} = \dfrac{1}{\sqrt6}\hat{i}+\dfrac{1}{\sqrt6}\hat{j}+\dfrac{2}{\sqrt6}\hat{k}.

\vec{PQ} = (4-1)\hat{i}+(5-2)\hat{j}+(6-3)\hat{k} = 3\hat{i}+3\hat{j}+3\hat{k}.

|\vec{PQ}| = \sqrt{9+9+9} = 3\sqrt3.

\widehat{PQ} = \dfrac{3\hat{i}+3\hat{j}+3\hat{k}}{3\sqrt3} = \dfrac{1}{\sqrt3}\hat{i}+\dfrac{1}{\sqrt3}\hat{j}+\dfrac{1}{\sqrt3}\hat{k}.

\vec{a}+\vec{b} = (2-1)\hat{i}+(-1+1)\hat{j}+(2-1)\hat{k} = \hat{i}+\hat{k}.

|\vec{a}+\vec{b}| = \sqrt{1^2+0^2+1^2} = \sqrt2.

Required unit vector = \dfrac{1}{\sqrt2}\hat{i}+\dfrac{1}{\sqrt2}\hat{k}.

Let \vec{a}=5\hat{i}-\hat{j}+2\hat{k}. Then |\vec{a}| = \sqrt{25+1+4} = \sqrt{30}.

Unit vector in the direction of \vec{a}: \hat{a} = \dfrac{1}{\sqrt{30}}(5\hat{i}-\hat{j}+2\hat{k}).

The required vector of magnitude 8 is 8\hat{a} = \dfrac{8}{\sqrt{30}}(5\hat{i}-\hat{j}+2\hat{k}) = \dfrac{40}{\sqrt{30}}\hat{i}-\dfrac{8}{\sqrt{30}}\hat{j}+\dfrac{16}{\sqrt{30}}\hat{k}.

Let \vec{a}=2\hat{i}-3\hat{j}+4\hat{k} and \vec{b}=-4\hat{i}+6\hat{j}-8\hat{k}.

Observe that \vec{b} = -2(2\hat{i}-3\hat{j}+4\hat{k}) = -2\vec{a}.

Since \vec{b}=\lambda\vec{a} for the scalar \lambda=-2, the two vectors are parallel to the same line.

Magnitude = \sqrt{1^2+2^2+3^2} = \sqrt{14}.

Direction cosines = \left(\dfrac{1}{\sqrt{14}}, \dfrac{2}{\sqrt{14}}, \dfrac{3}{\sqrt{14}}\right).

\vec{AB} = (-1-1)\hat{i}+(-2-2)\hat{j}+(1-(-3))\hat{k} = -2\hat{i}-4\hat{j}+4\hat{k}.

|\vec{AB}| = \sqrt{4+16+16} = \sqrt{36}=6.

Direction cosines = \left(-\dfrac{2}{6}, -\dfrac{4}{6}, \dfrac{4}{6}\right) = \left(-\dfrac13, -\dfrac23, \dfrac23\right).

Let \vec{a}=\hat{i}+\hat{j}+\hat{k}. Then |\vec{a}|=\sqrt{1+1+1}=\sqrt3.

Direction cosines: l=m=n=\dfrac{1}{\sqrt3}.

Since \cos\alpha=\cos\beta=\cos\gamma=\dfrac{1}{\sqrt3}, the angles α, β and γ that the vector makes with OX, OY and OZ are all equal.

Let \vec{a}=\vec{OP}=\hat{i}+2\hat{j}-\hat{k} and \vec{b}=\vec{OQ}=-\hat{i}+\hat{j}+\hat{k}, dividing in the ratio m : n = 2 : 1.

(i) Internally: \vec{OR} = \dfrac{m\vec{b}+n\vec{a}}{m+n} = \dfrac{2(-\hat{i}+\hat{j}+\hat{k})+1(\hat{i}+2\hat{j}-\hat{k})}{3}.

= \dfrac{(-2+1)\hat{i}+(2+2)\hat{j}+(2-1)\hat{k}}{3} = \dfrac{-\hat{i}+4\hat{j}+\hat{k}}{3}.

(ii) Externally: \vec{OR} = \dfrac{m\vec{b}-n\vec{a}}{m-n} = \dfrac{2(-\hat{i}+\hat{j}+\hat{k})-1(\hat{i}+2\hat{j}-\hat{k})}{1}.

= (-2-1)\hat{i}+(2-2)\hat{j}+(2+1)\hat{k} = -3\hat{i}+3\hat{k}.

The midpoint R of P and Q has position vector \vec{OR} = \dfrac{\vec{a}+\vec{b}}{2}, where \vec{a}=2\hat{i}+3\hat{j}+4\hat{k} and \vec{b}=4\hat{i}+\hat{j}-2\hat{k}.

\vec{OR} = \dfrac{(2+4)\hat{i}+(3+1)\hat{j}+(4-2)\hat{k}}{2} = \dfrac{6\hat{i}+4\hat{j}+2\hat{k}}{2}.

\vec{AB} = \vec{b}-\vec{a} = -\hat{i}+3\hat{j}+5\hat{k}, so |\vec{AB}|^2 = 1+9+25=35.

\vec{BC} = \vec{c}-\vec{b} = -\hat{i}-2\hat{j}-6\hat{k}, so |\vec{BC}|^2 = 1+4+36=41.

\vec{CA} = \vec{a}-\vec{c} = 2\hat{i}-\hat{j}+\hat{k}, so |\vec{CA}|^2 = 4+1+1=6.

Since |\vec{AB}|^2+|\vec{CA}|^2 = 35+6 = 41 = |\vec{BC}|^2, the converse of the Pythagoras theorem is satisfied.

By the triangle law, for any triangle ABC taken in order: \vec{AB}+\vec{BC}+\vec{CA}=\vec{0} — so (A) is true.

From this, \vec{AB}+\vec{BC}=-\vec{CA}=\vec{AC}, so \vec{AB}+\vec{BC}-\vec{AC}=\vec{0} — so (B) is true.

For (C), \vec{AB}+\vec{BC}-\vec{CA} = \vec{AC}-\vec{CA} = \vec{AC}+\vec{AC} = 2\vec{AC} \neq \vec{0} in general — so (C) is not true.

For (D), \vec{AB}-\vec{CB}+\vec{CA} = \vec{AB}+\vec{BC}+\vec{CA} = \vec{0} (using -\vec{CB}=\vec{BC}) — so (D) is true.

(A) is a correct statement — this is precisely the defining property of collinear vectors.

(C) is incorrect — for collinear vectors, the respective components are always proportional (that is exactly how collinearity is tested using components); the statement claims the opposite.

(D) is incorrect as a general statement — collinear vectors can point in opposite directions too (e.g. \vec{a} and -2\vec{a} are collinear but not in the same direction), so it is not true for every pair of collinear vectors.