Let `veca,vecb and vecc` be the non zero vectors such that `(vecaxxvecb)xxvecc=1/3 |vecb||vecc|veca.` if theta is the acute angle between the vectors

Question

Let `veca,vecb and vecc` be the non zero vectors such that `(vecaxxvecb)xxvecc=1/3 |vecb||vecc|veca.` if theta is the acute angle between the vectors

Let `veca,vecb and vecc` be the non zero vectors such that `(vecaxxvecb)xxvecc=1/3 |vecb||vecc|veca.` if theta is the acute angle between the vectors `vecb and veca` then theta equals (A) `1/3` (B) `sqrt(2)/3` (C) `2/3` (D) `2sqrt(2)/3`

Monjur Alahi

4 views

2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

we have `(veca xx vecb ) xx vecc = 1/3 |vecb||vecc|veca`
`or (veca .vecc) vecb - (vecb .vecc)veca = 1/3 |vecb||vecc|veca`
`or (veca. Vecc) vecb- {(vecb.vecc) + 1/3 |vecb|vecc|} veca =vec0`
`Rightarrow veca.vecc =0 and vecb.vecc + 1/3 |vecb|vecc|=0`
(`veca and vecb` are non-collinear)
or `|vecb||vecc|cos theta+ 1/3 |vecb |vecc| =0`
`or cos theta = -1//3`
`Rightarrow sin theta = sqrt(8/9) = ( 2sqrt2)/3`