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प्रश्न
If `veca = hati + hatj + hatk` and `vecb = hati + 2hatj + 3hatk` then find a unit vector perpendicular to both `veca + vecb` and `veca - vecb`.
उत्तर
`veca = hati + hatj + hatk` and `vecb = hati + 2hatj + 3hatk`
Now `veca + vecb = (hati + hatj + hatk) + (hati + 2hatj + 3hatk)`
= `2hati + 3hatj + 4hatk`
and `veca - vecb = (hati + hatj + hatk) - (hati + 2hatj + 3hatk)`
= `-hatj - 2hatk`
So `(veca + vecb) xx (veca - vecb)`
= `|(hati, hatj, hatk),(2, 3, 4),(0, -1, -2)|`
= `hati(-6 + 4) - hatj(-4) + hatk(-2)`
= `-2hati + 4hatj - 2hatk`
∴ Unit vector perpendicular to `(veca + vecb)` and `(veca - vecb)`.
= `(-2hati + 4hatj - 2hatk)/sqrt(4 + 16 + 4)`
= `(-2hati + 4hatj - 2hatk)/(2sqrt(6))`
= `(-hati + 2hatj - hatk)/sqrt(6)`
Therefore, unit vector perpendicular to `(veca + vecb)` and `(veca - vecb)` is `1/sqrt(6) (-hati + 2hatj - hatk)`.
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