If r- be a vector perpendicular to a-b-c- where -a- b-c-z and r- b-c-mc-a-na-b- then find l-m-n-

Given that #160; #160; r⃗ a⃗ + b⃗ + c⃗ So r⃗. (a⃗ + b⃗ + c⃗ +)=0 ⇒r⃗. a⃗+r⃗.b⃗+r⃗.c⃗ = 0 ......... 1 and z=[a⃗ b⃗&#16 ...

148

You visited us 148 times! Enjoying our articles? Unlock Full Access!

Linear Combination of Vectors

If r⃗ be a ...

Question

If $\to r$ be a vector perpendicular to $\to a + \to b + \to c$ , where $[\to a \to b \to c] = z$ and $\to r = ℓ (\to b \times \to c) + m (\to c \times \to a) + n (\to a \times \to b)$ , then find $l + m + n$ .

Open in App

Solution

Suggest Corrections

Similar questions

\to r

\to a + \to b + \to c

[\begin{matrix} \to a & \to b & \to c \end{matrix}] = 2

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

l + m + n

\to a, \to b

\to c

\to p, \to q

\to r

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

(\to a + \to b) \cdot \to p + (\to b + \to c) \cdot \to q + (\to c + \to a) \cdot \to r =

\to A = 2 \to i + \to k, \to B = \to i + \to j + \to k

\to C = 4 \to i - 3 \to j + 7 \to k

\to R

\to R \times \to B = \to C \times \to B

\to R \cdot \to A = 0

[\to a \times \to b \to b \times \to c \to c \times \to a] = λ [\to a \to b \to c]^{2}

λ

\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)

Join BYJU'S Learning Program