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1.

If ` vec a , vec b , vec c`are three mutually perpendicular vectors of equalmagniltgude, prove that ` vec a+ vec b+ vec c`is equally inclined with vectors ` vec a , vec b , a n d vec cdot`also find the angle.

Answer» `abs(veca)=abs(vecb)=abs(vecc)=lamda`
`veca*vecb=vecb*vecc*vecc*veca=0`
`abs(veca+vecb+vecc)^2=(veca+vecb+vecc)*(veca+vecb+vecc)`
`=abs(veca)^2+0+0+abs(vecb)^2+abs(vecc)^2`
`=lambda^2+lambda^2+lambda^2`
`=3lambda^2`
`abs(veca+vecb+vecc)=sqrt3lambda`
`costheta_1=(veca*(veca+vecb+vecc))/(abs(veca)abs(veca+vecb+vecc))=lambda^2/(lambdasqrt3lambda)=1/sqrt3=theta_1=cos^(-1)(1/sqrt3)`
Similarly,
`theta_1=theta_2=theta_3=cos^(-1)(1/sqrt3)`
Discussion
2.

If ` vec a`and ` vec b`are two vectors such that `| vec a|=4,| vec b|=3`and ` vec adot vec b=6`. Find the angle between ` vec a`and ` vec b`.

Answer» We know, `veca*vecb = |veca|*|vecb|cos theta`
Putting given values,
`6 = 4*3*costheta`
`=>cos theta = 1/2`
`=>theta = 60^@`
So, the angle between `veca` and `vecb` is `60^@`.
Discussion
3.

Find the angles of a triangle whose verticesare `A(0,-1,-2),B(3,1,4)`and `C`(5,7,1).

Answer» Here, `vec(AB) = vecB - vecA = 3hati+2hatj+6hatk`
`vec(BC) = 2hati+6hatj-3hatk`
`vec(AC) = 5hati+8hatj+3hatk`
`|vec(AB)| = sqrt(3^2+2^2+6^2) = 7`
`|vec(BC)| = sqrt(32^2+6^2+(-3)^2) = 7`
`|vec(AC)| = sqrt(5^2+8^2+3^2) = 7sqrt2`
Now, `vec(AB)*vec(BC) = 6+12-18 = 0`
It means, `AB` and `BC` are perpendicular.
`:./_ABC = 90^@`
Also, as `|vec(AB)| = |vec(BC)| `
`:. /_ACB = /_BAC`
As, `/_ABC = 90^@`
`:. /_ACB = /_BAC = 1/2*(180-90)^@ = 45^@.`
Discussion
4.

If ` hat a`and ` hat b`are unit vectors inclined at an angle `theta`, then prove that`costheta/2=1/2| hat a+ hat b|``tantheta/2=1/2|( hat a- hat b)/( hat a+ hat b)|`

Answer» Here, `veca` and `vecb` are unit vectors.
`:. |hata| = |hatb| = 1`
(i) `|hata+hatb|^2 = (hata)^2+(hatb)^2+2hata*hatb`
`=|hata|^2+|hatb|^2+2|hata||hatb|cos theta`
`=1^2+1^2+2(1)(1)costheta`
`=>|hata+hatb|^2=2(1+costheta)`
`=>|hata+hatb|^2/2 = 1+costheta`
`=>|hata+hatb|^2/2 = 2cos^2(theta/2)`
`=>|hata+hatb|^2/4 = cos^2(theta/2)`
`=>cos (theta/2) =1/2 |hata+hatb| ->(1)`

(ii) `|hata-hatb|^2 = (hata)^2+(hatb)^2-2hata*hatb`
`=|hata|^2+|hatb|^2-2|hata||hatb|cos theta`
`=1^2+1^2-2(1)(1)costheta`
`=>|hata-hatb|^2=2(1-costheta)`
`=>|hata-hatb|^2/2 = 1-costheta`
`=>|hata-hatb|^2/2 = 2sin^2(theta/2)`
`=>|hata-hatb|^2/4 = sin^2(theta/2)`
`=>sin (theta/2) =1/2 |hata-hatb| ->(2)`
Now, dividing (2) by (1),
`sin (theta/2)/cos (theta/2) = |hata-hatb|/|hata+hatb|`
`=> tan (theta/2) = |hata-hatb|/|hata+hatb|`
Discussion