Solution : We know that the vector equation of line passing through two points with position vectors \(\vec{a}\) and \(\vec{b}\) is, \(\vec{r}\) = \(\lambda\) \((\vec{b} – \vec{a})\) Here \(\vec{a}\) = \(3\hat{i} + 4\hat{j} – 7\hat{k}\) and \(\vec{b}\) = \(\hat{i} – \hat{j} + 6\hat{k}\). So, the vector equation of the required line is \(\vec{r}\) = (\(3\hat{i} […]
Solution : We have, \(\vec{a}\) = \(4\hat{i} – 3\hat{j} + 5\hat{k}\) and \(\vec{b}\) = \(3\hat{i} + 4\hat{j} + 5\hat{k}\) Let \(\theta\) is the angle between the given vectors. Then, cos\(\theta\) = \(\vec{a}.\vec{b}\over |\vec{a}||\vec{b}|\) \(\implies\) cos\(\theta\) = \(12 – 12 + 25\over \sqrt{16 + 9 + 25} \sqrt{16 + 9 + 25}\) = \(1\over 2\) \(\implies\)
Solution : We have \(\vec{a}\) = \(2\hat{i}+2\hat{j}-\hat{k}\) and \(\vec{b}\) = \(6\hat{i}-3\hat{j}+2\hat{k}\) \(\vec{a}\).\(\vec{b}\) = (\(2\hat{i}+2\hat{j}-\hat{k}\)).(\(6\hat{i}-3\hat{j}+2\hat{k}\)) = (2)(6) + (2)(-3) + (-1)(2) = 12 – 6 – 2 = 4 Similar Questions Find the angle between the vectors with the direction ratios proportional to 4, -3, 5 and 3, 4, 5. Find the vector equation of a
Solution : We have [\(\vec{a}\) + \(\vec{b}\) \(\vec{b}\) + \(\vec{c}\) \(\vec{c}\) + \(\vec{a}\)] = {(\(\vec{a}\) + \(\vec{b}\))\(\times\)(\(\vec{b}\) + \(\vec{c}\))}.(\(\vec{c}\) + \(\vec{a}\)) = {\(\vec{a}\)\(\times\)\(\vec{b}\) + \(\vec{a}\)\(\times\)\(\vec{c}\) + \(\vec{b}\)\(\times\)\(\vec{b}\) + \(\vec{b}\)\(\times\)\(\vec{c}\)}.(\(\vec{c}\) + \(\vec{a}\)) {\(\vec{b}\)\(\times\)\(\vec{b}\) = 0} = {\(\vec{a}\)\(\times\)\(\vec{b}\) + \(\vec{a}\)\(\times\)\(\vec{c}\) + \(\vec{b}\)\(\times\)\(\vec{c}\)}.(\(\vec{c}\) + \(\vec{a}\)) = (\(\vec{a}\times\vec{b}\)).\(\vec{c}\) + (\(\vec{a}\times\vec{c}\)).\(\vec{c}\) + (\(\vec{b}\times\vec{c}\)).\(\vec{c}\) + (\(\vec{a}\times\vec{b}\)).\(\vec{a}\) + (\(\vec{a}\times\vec{c}\)).\(\vec{a}\) + (\(\vec{b}\times\vec{c}\)).\(\vec{a}\) =
Solution : \(\vec{a}\times\vec{b}\) = \(\vec{c}\) and \(\vec{b}\times\vec{c}\) = \(\vec{a}\) \(\implies\) \(\vec{c}\perp\vec{a}\) , \(\vec{c}\perp\vec{b}\) and \(\vec{a}\perp\vec{b}\), \(\vec{a}\perp\vec{c}\) \(\implies\) \(\vec{a}\perp\vec{b}\), \(\vec{b}\perp\vec{c}\) and \(\vec{c}\perp\vec{a}\) \(\implies\) \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are mutually perpendicular vectors. Again, \(\vec{a}\times\vec{b}\) = \(\vec{c}\) and \(\vec{b}\times\vec{c}\) = \(\vec{a}\) \(\implies\) |\(\vec{a}\times\vec{b}\)| = |\(\vec{c}\)| and |\(\vec{b}\times\vec{c}\)| = |\(\vec{a}\)| \(\implies\) \(|\vec{a}||\vec{b}|sin{\pi\over 2}\) = |\(\vec{c}\)| and \(|\vec{b}||\vec{c}|sin{\pi\over 2}\) = |\(\vec{a}\)|
Solution : Unit vectors perpendicular to \(\vec{a}\) & \(\vec{b}\) = \(\pm\)\(\vec{a}\times\vec{b}\over |\vec{a}\times\vec{b}|\) \(\therefore\) \(\vec{a}\times\vec{b}\) = \(\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -3 \\ 1 & 2 & -2 \\ \end{vmatrix}\) = \(-5\hat{i} – 5\hat{j} – 5\hat{k}\) \(\therefore\) Unit Vectors = \(\pm\) \(-5\hat{i} – 5\hat{j} – 5\hat{k}\over 5\sqrt{3}\) Hence the required
