Question

Find the unit vector perpendicular to Ā= 31+2) - and B=1-1+1​

Answer 1

Finally, the required, unit vector will be

∣∣AB×AC∣∣AB×AC

We have, AB=(1−3,−1+1,−3−2)=(−2,0,−5).

$AC=(1,−2,−1), so that,</u></strong></p><p></p><p><strong><u>[tex]AC=(1,−2,−1), so that,AB×AC=∣∣∣∣∣∣∣∣i−21j0−2k−5−1∣∣∣∣∣∣∣∣</u></strong></p><p></p><p></p><p><strong><u>[tex]AC=(1,−2,−1), so that,AB×AC=∣∣∣∣∣∣∣∣i−21j0−2k−5−1∣∣∣∣∣∣∣∣$

=−10i−7j+4k=(−10,−7,4)

=−10i−7j+4k=(−10,−7,4)⇒∣∣AB×AC∣∣=(−10)2+(−7)2+(4)2=100+49+16=165.

=−10i−7j+4k=(−10,−7,4)⇒∣∣AB×AC∣∣=(−10)2+(−7)2+(4)2=100+49+16=165.Finally, the desired unit vector is 

=−10i−7j+4k=(−10,−7,4)⇒∣∣AB×AC∣∣=(−10)2+(−7)2+(4)2=100+49+16=165.Finally, the desired unit vector is (−16510,−1657,1654).

=−10i−7j+4k=(−10,−7,4)⇒∣∣AB×AC∣∣=(−10)2+(−7)2+(4)2=100+49+16=165.Finally, the desired unit vector is (−16510,−1657,1654).

Answer 2

ANSWER ➡️The given points, A(3,−1,2),B(1,−1,−3) and C(4,−3,1) lie in the plane ABC.

Accordingly, the vectors

AB

and

AC

ϵ the plane ABC.

Hence,

AB

×

AC

is perpendicular to the plane ABC.

Finally, the required, unit vector will be

∣∣

AB

×

AC

∣∣

AB

×

AC

We have,

AB

=(1−3,−1+1,−3−2)=(−2,0,−5).

AC

=(1,−2,−1), so that,

AB

×

AC

=

∣

∣

∣

∣

∣

∣

∣

∣

i

−2

1

j

0

−2

k

−5

−1

∣

∣

∣

∣

∣

∣

∣

∣

=−10i−7j+4k=(−10,−7,4)

⇒∣∣

AB

×

AC

∣∣=

(−10)

2

+(−7)

2

+(4)

2

=

100+49+16

=

165

.

Finally, the desired unit vector is

(− 16510,−1657,1654).