Let r- be a vector perpendicular to a-b-c- where - a- b- c- -2- If r-l b- c -m c- a -n a- b then l-m-n is

The correct option is C 0 r⃗ be a vector perpendicular to a⃗+b⃗+c⃗ , where [ a⃗ amp; #160; b⃗ amp; c⃗ ]=2 i.e r⃗·(a⃗+b⃗+c⃗ ...

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Determinant

Let r⃗ be a...

Question

Let $\to r$ be a vector perpendicular to $\to a + \to b + \to c$ , where $[\begin{matrix} \to a & \to b & \to c \end{matrix}] = 2$ . If $\to r = l (b \times c) + m (c \times a) + n (a \times b)$ then $l + m + n$ is

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\to r

\to a + \to b + \to c

[\to a \to b \to c] = 2

\to r = l (\to b \times \to c) + m (\to c \times \to a) + n (\to a \times \to b)

\to r

\to a + \to b + \to c

[\to a \to b \to c] = z

\to r = ℓ (\to b \times \to c) + m (\to c \times \to a) + n (\to a \times \to b)

l + m + n

\to r = (\to a \times \to b) sin x + (\to b \times \to c) cos y + 2 (\to c \times \to a)

\to a \to b \to c

\to r

\to a + \to b + \to c

x^{2} + y^{2}

¯ ¯ ¯ a, ¯ ¯ b, ¯ ¯ c

¯ ¯ ¯ p, ¯ ¯ ¯ q, ¯ ¯ ¯ r

\to p = \frac{\to b \times \to c}{[\begin{matrix} \to a & \to b & \to c \end{matrix}]}, \to q = \frac{\to c \times \to a}{[\begin{matrix} \to a & \to b & \to c \end{matrix}]}, \to r = \frac{\to a \times \to b}{[\begin{matrix} \to a & \to b & \to c \end{matrix}]}

(\to a + \to b + \to c) . (\to p + \to q + \to r)

\to a, \to b, \to c

\to r

(\to a \times \to b) \times (\to r \times \to c) + (\to b \times \to c) \times (\to r \times \to a) + (\to c \times \to a) \times (\to r \times \to b)

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