Let a-b and c be mutually perpendicular vectors of the same magnitude-If a vector x satisfies the equation a-x-b-a-b-x-c-b-c-x-a-c-0- then x is given by

The correct option is A 12(a+b+c) Let |a|=|b|=|c|=λ , then nbsp; a^2(x b) [a.(x b)]+b^2(x c) [b.(x c)]b+c^2(x a) c(x a]c=0 λ^2(3x a b c) [ ...

You visited us 18 times! Enjoying our articles? Unlock Full Access!

Convexity

Let a,b and...

Question

Let $\to a, \to b$ and $\to c$ be mutually perpendicular vectors of the same magnitude.If a vector $\to x$ satisfies the equation
$\to a \times [(\to x - \to b) \times \to a] + \to b \times [(\to x - \to c) \times \to b] + \to c \times [(\to x - \to a) \times \to c] = 0,$ then $\to x$ is given by

\frac{1}{2} (\to a + \to b + \to c)

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{1}{3} (2 \to a + \to b + \to c)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1}{3} (\to a + \to b + \to c)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{1}{2} (\to a + \to b - 2 \to c)

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

Suggest Corrections

Similar questions

\to a =^i +^j +^k, \to b =^i -^j + 2^k

\to c = x^i + (x - 2)^j -^k

\to c

\to a

\to b

x =

a, b, c

a \neq 0

a \times b = 2 a \times c, | a | = | c | = 1,

| b | = 4

| b \times c | = \sqrt{15}

b - 2 c = λ a

λ

¯ ¯ ¯ a, ¯ ¯ b, ¯ ¯ c

3 ¯ ¯ ¯ a - 2 ¯ ¯ b - 4 ¯ ¯ c

- ¯ ¯ ¯ a + 2 ¯ ¯ c

- 2 ¯ ¯ ¯ a + ¯ ¯ b + 3 ¯ ¯ c

\to a, \to b, \to c

\to r

\to a \times {(\to r - \to b) \times \to a} + \to b \times {(\to r - \to c) \times \to b} + \to c \times {(\to r - \to a) \times \to c} = \to 0

\to r

A = (1, 1, 1), C = (0, 1, - 1)

B

A \times B = C

A \cdot B = 3

9 {| B |}^{2}

Join BYJU'S Learning Program