MHT CET · Maths · Vector Algebra
If \(|\overline{\mathrm{a}}|=2,|\overline{\mathrm{b}}|=3,|\overline{\mathrm{c}}|=5\) and each of the angles between the vectors \(\bar{a}\) and \(\bar{b}, \bar{b}\) and \(\bar{c}\), \(\overline{\mathrm{c}}\) and \(\overline{\mathrm{a}}\) is \(60^{\circ}\), then the value of \(|\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}|\) is
- A \(\sqrt {69}\)
- B \(\sqrt {70}\)
- C \(\sqrt {80}\)
- D \(\sqrt {39}\)
Answer & Solution
Correct Answer
(A) \(\sqrt {69}\)
Step-by-step Solution
Detailed explanation
\(\overline{\mathrm{a} \cdot \overline{\mathrm{b}}} =|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}| \cos 60^{\circ} \)
\( =(2)(3)\left(\frac{1}{2}\right) \)
\( =3 \)
\( \overline{\mathrm{b}} \cdot \overline{\mathrm{c}} =|\overline{\mathrm{b}}||\overline{\mathrm{c}}| \cos 60^{\circ} \)
\( =(3)(5)\left(\frac{1}{2}\right)=\frac{15}{2}\)
\(\overline{\mathrm{a} \cdot \overline{\mathrm{c}}} =|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{c}}| \cos 60^{\circ} \)
\( =(2)(5)\left(\frac{1}{2}\right) \)
\( =5 \)
\( \therefore \mid \overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}^2 =\left.\left|\overline{\mathrm{a}}^2+\right| \overrightarrow{\mathrm{b}}\right|^2+|\overline{\mathrm{c}}|^2~+\) \(2(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}+\overline{\mathrm{b}} \cdot \overline{\mathrm{c}}+\overline{\mathrm{c}} \cdot \overline{\mathrm{a}}) \)
\( =2^2+3^2+5^2+2\left(3+\frac{15}{2}+5\right) \)
\( =4+9+25+31 \)
\( =69 \)
\( \therefore \quad|\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}| =\sqrt{69}\)
\( =(2)(3)\left(\frac{1}{2}\right) \)
\( =3 \)
\( \overline{\mathrm{b}} \cdot \overline{\mathrm{c}} =|\overline{\mathrm{b}}||\overline{\mathrm{c}}| \cos 60^{\circ} \)
\( =(3)(5)\left(\frac{1}{2}\right)=\frac{15}{2}\)
\(\overline{\mathrm{a} \cdot \overline{\mathrm{c}}} =|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{c}}| \cos 60^{\circ} \)
\( =(2)(5)\left(\frac{1}{2}\right) \)
\( =5 \)
\( \therefore \mid \overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}^2 =\left.\left|\overline{\mathrm{a}}^2+\right| \overrightarrow{\mathrm{b}}\right|^2+|\overline{\mathrm{c}}|^2~+\) \(2(\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}+\overline{\mathrm{b}} \cdot \overline{\mathrm{c}}+\overline{\mathrm{c}} \cdot \overline{\mathrm{a}}) \)
\( =2^2+3^2+5^2+2\left(3+\frac{15}{2}+5\right) \)
\( =4+9+25+31 \)
\( =69 \)
\( \therefore \quad|\overline{\mathrm{a}}+\overline{\mathrm{b}}+\overline{\mathrm{c}}| =\sqrt{69}\)
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