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Show that (vecaxxvecb)^2 = a^2b^2 - (veca.vecb)^2. |
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Answer» SOLUTION :`vecaxxvecb = ab sinthetahatn` Where a = `|veca|, B = |vecb|` = `THETA` is the angle between two vectors `veca` and `vecb` and `hatn` is the unit vector perpendicular to `veca.vecb`. Now `(vecaxxvecb)^2 = (vecaxxvecb).(vecaxxvecb) = `a^2b^2 sin^2 theta (hatn.hatn)` = `a^2 b^2 sin^2 theta = a^2b^2- a^2b^2 cos^2 theta`. = `a^2b^2-(abcostheta)^2 = a^2b^2-(veca.vecb)^2`. (PROVED) |
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