If-a-i-j-k-b-i-j-k-and-c-is-a-unit-vector-to-the-vector-a-and-coplanar-with-a-and-b-then-a-unit-vector-d-to-both-a-and-c-is-

Question Number 54419 by gunawan last updated on 03/Feb/19

If a = i + j - k, b = i - j + k and c is a unit

vector ⊥ to the vector a and coplanar

with a and b, then a unit vector d ⊥ to

both a and c is

Answered by tanmay.chaudhury50@gmail.com last updated on 03/Feb/19

c = x a + y b

c = x (i + j - k) + y (i - j + k)

c = i (x + y) + j (x - y) + k (- x + y)

c . a = 0

[i (x + y) + j (x - y) + k (- x + y)] . [i + j - k] = 0

(x + y) + (x - y) + (- x + y) (- 1) = 0

x + y + x - y + x - y = 0

3 x - y = 0 \dots . . (1) e q n

c = i (x + y) + j (x - y) + k (- x + y)

c = i (x + 3 x) + j (x - 3 x) + k (- x + 3 x)

c = i (4 x) + j (- 2 x) + k (2 x)

g i v e n \sqrt{16 x^{2} + 4 x^{2} + 4 x^{2}} = 1

\sqrt{24 x^{2}} = 1 x = \pm \frac{1}{\sqrt{24}}

c o n s i d e r i n g + v e v a l u e o n l y \dots x = \frac{1}{\sqrt{24}} y = \frac{3}{\sqrt{24}}

c = i (x + y) + j (x - y) + k (- x + y)

c = i (\frac{4}{\sqrt{24}}) + j (\frac{- 2}{\sqrt{24}}) + k (\frac{2}{\sqrt{24}})

c o n s i d e r i n g - v e s i g n \dots

c = i (\frac{- 4}{\sqrt{24}}) + j (\frac{2}{\sqrt{24}}) + k (\frac{- 2}{\sqrt{24}})

d = \frac{a \times c}{∣ a \times c ∣} n o w a \times c

[i j k]

[1 1 - 1]

[\frac{4}{\sqrt{24}} \frac{- 2}{\sqrt{24}} \frac{2}{\sqrt{24}}]

n o w y o u s o l v e t h e r e m a i n i n g \dots