Find-a-unit-vector-which-is-perpendicula-to-a-vector-A-3coma5coma1-Sorry-for-writing-coma-cuz-i-dnt-see-a-key-for-it-

Question Number 23442 by chernoaguero@gmail.com last updated on 30/Oct/17

Find a unit vector which is perpendicula

to a vector A (3 coma 5 coma 1)

Sorry for writing coma cuz i dnt see a key for it

Commented by $@ty@m last updated on 31/Oct/17

Commented by chernoaguero@gmail.com last updated on 31/Oct/17

oopz u are right i was seachin for it

since but could not see iti

Answered by $@ty@m last updated on 31/Oct/17

G i v e n,

\vec{O A} = 3 \hat{i} + 5 \hat{j} + \hat{k}

L e t \vec{O B} = x \hat{i} + y \hat{j} + z \hat{k} s u c h t h a t

\vec{O A} ⊥ \vec{O B} & ∣ O B ∣= 1 \dots . (1)

\Rightarrow \vec{O A} . \vec{O B} = 0

\Rightarrow 3 x + 5 y + z = 0 \dots . (2)

T h e r e w o u l d b e i n f i n i t e u n i t

v e c t o r s \vec{O B} = x \hat{i} + y \hat{j} + z \hat{k}

s a t i s f y i n g c o n d i t i o n s (1) a n d (2) .

F o r e x a m p l e,

L e t x = 0

t h e n y^{2} + z^{2} = 1 & 5 y + z = 0

\Rightarrow y = \pm \frac{\sqrt{26}}{26} & z = \mp \frac{5 \sqrt{26}}{26}

∴ o n e o f t h e p o s s i b l e u n i t v e c t o r s

p e r p e n d i c u l a r t o \vec{O A} i s

\frac{\sqrt{26}}{26} \hat{j} - \frac{5 \sqrt{26}}{26} \hat{k}

Commented by chernoaguero@gmail.com last updated on 01/Nov/17

Thank u so much