Find dot product of vectors \(\vec{a}\) = \(2\hat{i}+2\hat{j}-\hat{k}\) and \(\vec{b}\) = \(6\hat{i}-3\hat{j}+2\hat{k}\)

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 line which passes through the point A (3, 4, -7) and B (1, -1, 6)

For any three vectors \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) prove that [\(\vec{a}\) + \(\vec{b}\) \(\vec{b}\) + \(\vec{c}\) \(\vec{c}\) + \(\vec{a}\)] = 2[\(\vec{a}\) \(\vec{b}\) \(\vec{c}\)]

If \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are three non zero vectors such that \(\vec{a}\times\vec{b}\) = \(\vec{c}\) and \(\vec{b}\times\vec{c}\) = \(\vec{a}\), prove that \(\vec{a}\), \(\vec{b}\), \(\vec{c}\) are mutually at right angles and |\(\vec{b}\)| = 1 and |\(\vec{c}\)| = |\(\vec{a}\)|

Find the vector of magnitude 5 which are perpendicular to the vectors \(\vec{a}\) = \(2\hat{i} + \hat{j} – 3\hat{k}\) and \(\vec{b}\) = \(\hat{i} – 2\hat{j} + \hat{k}\)

Leave a Comment

Your email address will not be published.