NCERT Solutions · Class 12 Mathematics Vector Algebra

Two nonzero vectors are perpendicular if and only if their dot product is zero, so compute the dot product of the two given vectors.Using distributivity, and writing \(\displaystyle |\vec{a}|,|\vec{b}|\) as scalars that may be moved out: \[\big(|\vec{a}|\vec{b}+|\vec{b}|\vec{a}\big)\cdot\big(|\vec{a}|\vec{b}-|\vec{b}|\vec{a}\big)\] \[=|\vec{a}|^{2}(\vec{b}\cdot\vec{b})-|\vec{a}||\vec{b}|(\vec{b}\cdot\vec{a})+|\vec{b}||\vec{a}|(\vec{a}\cdot\vec{b})-|\vec{b}|^{2}(\vec{a}\cdot\vec{a})\] The two middle terms cancel because \(\displaystyle \vec{b}\cdot\vec{a}=\vec{a}\cdot\vec{b}\), and \(\displaystyle \vec{b}\cdot\vec{b}=|\vec{b}|^{2}\), \(\displaystyle \vec{a}\cdot\vec{a}=|\vec{a}|^{2}\): \[=|\vec{a}|^{2}|\vec{b}|^{2}-|\vec{b}|^{2}|\vec{a}|^{2}=0\] Since \(\displaystyle \vec{a}\) and \(\displaystyle \vec{b}\) are nonzero, the dot product being \(\displaystyle 0\) shows that \(\displaystyle |\vec{a}|\vec{b}+|\vec{b}|\vec{a}\) is perpendicular to \(\displaystyle |\vec{a}|\vec{b}-|\vec{b}|\vec{a}\).

Use \(\displaystyle \vec{a}\cdot\vec{a}=|\vec{a}|^{2}\). The first condition gives \[|\vec{a}|^{2}=0\ \Rightarrow\ |\vec{a}|=0\ \Rightarrow\ \vec{a}=\vec{0}\] So \(\displaystyle \vec{a}\) is forced to be the zero vector.But the zero vector has zero dot product with every vector: \(\displaystyle \vec{0}\cdot\vec{b}=0\) for all \(\displaystyle \vec{b}\). Hence the second condition \(\displaystyle \vec{a}\cdot\vec{b}=0\) is automatically satisfied and imposes no restriction whatever on \(\displaystyle \vec{b}\).Conclusion: nothing can be concluded about \(\displaystyle \vec{b}\) — it may be any vector at all.

The standard move is to take the square of the modulus of the given vector equation, using \(\displaystyle |\vec{v}|^{2}=\vec{v}\cdot\vec{v}\).From \(\displaystyle \vec{a}+\vec{b}+\vec{c}=\vec{0}\), \[|\vec{a}+\vec{b}+\vec{c}|^{2}=0\] \[(\vec{a}+\vec{b}+\vec{c})\cdot(\vec{a}+\vec{b}+\vec{c})=0\] Expanding, and pairing terms with commutativity of the dot product, \[|\vec{a}|^{2}+|\vec{b}|^{2}+|\vec{c}|^{2}+2\big(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a}\big)=0\] Since \(\displaystyle \vec{a},\vec{b},\vec{c}\) are unit vectors, \(\displaystyle |\vec{a}|^{2}=|\vec{b}|^{2}=|\vec{c}|^{2}=1\): \[3+2\big(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a}\big)=0\] \[\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a}=-\frac{3}{2}\]

The converse would say: if \(\displaystyle \vec{a}\cdot\vec{b}=0\) then \(\displaystyle \vec{a}=\vec{0}\) or \(\displaystyle \vec{b}=\vec{0}\). This is false, because \(\displaystyle \vec{a}\cdot\vec{b}=|\vec{a}||\vec{b}|\cos\theta\) can vanish through the third factor \(\displaystyle \cos\theta\) — that is, when the vectors are perpendicular \(\displaystyle (\theta=90^{\circ})\) — without either magnitude being zero.Counter-example. Take \[\vec{a}=\hat{i}+\hat{j},\qquad \vec{b}=\hat{i}-\hat{j}\] Then \[|\vec{a}|=\sqrt{1^{2}+1^{2}}=\sqrt{2}\neq 0,\qquad |\vec{b}|=\sqrt{1^{2}+(-1)^{2}}=\sqrt{2}\neq 0\] so neither vector is the zero vector, yet \[\vec{a}\cdot\vec{b}=(1)(1)+(1)(-1)=1-1=0\] Hence \(\displaystyle \vec{a}\cdot\vec{b}=0\) does not force \(\displaystyle \vec{a}=\vec{0}\) or \(\displaystyle \vec{b}=\vec{0}\); the converse need not be true.

\(\displaystyle \angle\mathrm{ABC}\) is the angle at the vertex \(\displaystyle \mathrm{B}\), so both vectors must start at \(\displaystyle \mathrm{B}\): use \(\displaystyle \overrightarrow{\mathrm{BA}}\) and \(\displaystyle \overrightarrow{\mathrm{BC}}\) (not \(\displaystyle \overrightarrow{\mathrm{AB}}\), which would give the supplement of the answer).With \(\displaystyle \mathrm{A}(1,2,3)\), \(\displaystyle \mathrm{B}(-1,0,0)\), \(\displaystyle \mathrm{C}(0,1,2)\), \[\overrightarrow{\mathrm{BA}}=\big(1-(-1)\big)\hat{i}+(2-0)\hat{j}+(3-0)\hat{k}=2\hat{i}+2\hat{j}+3\hat{k}\] \[\overrightarrow{\mathrm{BC}}=\big(0-(-1)\big)\hat{i}+(1-0)\hat{j}+(2-0)\hat{k}=\hat{i}+\hat{j}+2\hat{k}\] Then \[\overrightarrow{\mathrm{BA}}\cdot\overrightarrow{\mathrm{BC}}=(2)(1)+(2)(1)+(3)(2)=2+2+6=10\] \[|\overrightarrow{\mathrm{BA}}|=\sqrt{4+4+9}=\sqrt{17},\qquad |\overrightarrow{\mathrm{BC}}|=\sqrt{1+1+4}=\sqrt{6}\] Using \(\displaystyle \cos\theta=\dfrac{\overrightarrow{\mathrm{BA}}\cdot\overrightarrow{\mathrm{BC}}}{|\overrightarrow{\mathrm{BA}}||\overrightarrow{\mathrm{BC}}|}\), \[\cos(\angle \mathrm{ABC})=\frac{10}{\sqrt{17}\cdot\sqrt{6}}=\frac{10}{\sqrt{102}}\] \[\angle \mathrm{ABC}=\cos^{-1}\!\left(\frac{10}{\sqrt{102}}\right)\]

Three points are collinear when the sum of two of the distances between them equals the third, i.e. \(\displaystyle |\overrightarrow{\mathrm{AB}}|+|\overrightarrow{\mathrm{BC}}|=|\overrightarrow{\mathrm{AC}}|\) (equivalently, when \(\displaystyle \overrightarrow{\mathrm{AB}}\) and \(\displaystyle \overrightarrow{\mathrm{BC}}\) are parallel with a common point).From \(\displaystyle \mathrm{A}(1,2,7)\), \(\displaystyle \mathrm{B}(2,6,3)\), \(\displaystyle \mathrm{C}(3,10,-1)\), \[\overrightarrow{\mathrm{AB}}=(2-1)\hat{i}+(6-2)\hat{j}+(3-7)\hat{k}=\hat{i}+4\hat{j}-4\hat{k}\] \[\overrightarrow{\mathrm{BC}}=(3-2)\hat{i}+(10-6)\hat{j}+(-1-3)\hat{k}=\hat{i}+4\hat{j}-4\hat{k}\] \[\overrightarrow{\mathrm{AC}}=(3-1)\hat{i}+(10-2)\hat{j}+(-1-7)\hat{k}=2\hat{i}+8\hat{j}-8\hat{k}\] Their magnitudes are \[|\overrightarrow{\mathrm{AB}}|=\sqrt{1+16+16}=\sqrt{33},\qquad |\overrightarrow{\mathrm{BC}}|=\sqrt{1+16+16}=\sqrt{33}\] \[|\overrightarrow{\mathrm{AC}}|=\sqrt{4+64+64}=\sqrt{132}=2\sqrt{33}\] Since \[|\overrightarrow{\mathrm{AB}}|+|\overrightarrow{\mathrm{BC}}|=\sqrt{33}+\sqrt{33}=2\sqrt{33}=|\overrightarrow{\mathrm{AC}}|\] the points \(\displaystyle \mathrm{A},\mathrm{B},\mathrm{C}\) are collinear. (Indeed \(\displaystyle \overrightarrow{\mathrm{AC}}=2\,\overrightarrow{\mathrm{AB}}\), so \(\displaystyle \mathrm{B}\) is the midpoint of \(\displaystyle \mathrm{AC}\).)

Let the three points \(\displaystyle \mathrm{A},\mathrm{B},\mathrm{C}\) have position vectors \[\vec{\mathrm{A}}=2\hat{i}-\hat{j}+\hat{k},\qquad \vec{\mathrm{B}}=\hat{i}-3\hat{j}-5\hat{k},\qquad \vec{\mathrm{C}}=3\hat{i}-4\hat{j}-4\hat{k}\] A triangle is right angled when one pair of sides meeting at a vertex has zero dot product; equivalently when Pythagoras' relation holds among the squared side lengths.Side vectors: \[\overrightarrow{\mathrm{AB}}=\vec{\mathrm{B}}-\vec{\mathrm{A}}=-\hat{i}-2\hat{j}-6\hat{k},\qquad |\overrightarrow{\mathrm{AB}}|^{2}=1+4+36=41\] \[\overrightarrow{\mathrm{BC}}=\vec{\mathrm{C}}-\vec{\mathrm{B}}=2\hat{i}-\hat{j}+\hat{k},\qquad |\overrightarrow{\mathrm{BC}}|^{2}=4+1+1=6\] \[\overrightarrow{\mathrm{CA}}=\vec{\mathrm{A}}-\vec{\mathrm{C}}=-\hat{i}+3\hat{j}+5\hat{k},\qquad |\overrightarrow{\mathrm{CA}}|^{2}=1+9+25=35\] All three lengths are nonzero and \[|\overrightarrow{\mathrm{BC}}|^{2}+|\overrightarrow{\mathrm{CA}}|^{2}=6+35=41=|\overrightarrow{\mathrm{AB}}|^{2}\] So the longest side \(\displaystyle \mathrm{AB}\) is the hypotenuse and the right angle is at the opposite vertex \(\displaystyle \mathrm{C}\). Confirm it directly with the dot product of the two sides meeting at \(\displaystyle \mathrm{C}\): \[\overrightarrow{\mathrm{CB}}=\vec{\mathrm{B}}-\vec{\mathrm{C}}=-2\hat{i}+\hat{j}-\hat{k},\qquad \overrightarrow{\mathrm{CA}}=-\hat{i}+3\hat{j}+5\hat{k}\] \[\overrightarrow{\mathrm{CB}}\cdot\overrightarrow{\mathrm{CA}}=(-2)(-1)+(1)(3)+(-1)(5)=2+3-5=0\] Hence the three points form the vertices of a triangle that is right angled at \(\displaystyle \mathrm{C}\).

For a scalar \(\displaystyle \lambda\), \(\displaystyle |\lambda\vec{a}|=|\lambda|\,|\vec{a}|\). The vector \(\displaystyle \lambda\vec{a}\) is a unit vector precisely when this magnitude equals \(\displaystyle 1\).With \(\displaystyle |\vec{a}|=a\), \[|\lambda\vec{a}|=1\ \Rightarrow\ |\lambda|\,a=1\ \Rightarrow\ a=\frac{1}{|\lambda|}\] (The division is legitimate since \(\displaystyle \lambda\neq 0\), so \(\displaystyle |\lambda|\neq 0\).) Note that \(\displaystyle |\lambda|\) must be used, not \(\displaystyle \lambda\): a negative \(\displaystyle \lambda\) reverses direction but cannot make a length negative.The correct option is (D) \(\displaystyle a=1/|\lambda|\).