Three vectors P- Q- R- are such that the -P- -Q- -R- -2 -P- and P-Q- -R-0- The angle between P- and Q- Q- and R- and P- and R- will be respectively-

The correct option is A 90^o, 135^o, 135^o SQUARING P+Q+R=0 gives P^2+Q^2+R^2+2PQ cosθ_1+2QR #10;cos θ_2+2PR cos θ_3=0 .Now putting the value ...

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Determinant

Three vectors...

Question

Three vectors $\to P$ , $\to Q$ , $\to R$ are such that the $| \to P |$ $= | \to Q |$ , $| \to R |$ $= \sqrt{2}$ $| \to P |$ and $\to P + \to Q$ $+ \to R = 0$ . The angle between $\to P$ and $\to Q$ , $\to Q$ and $\to R$ and $\to P$ and $\to R$ will be respectively.

90^{o}, 135^{o}, 135^{o}

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90^{o}, 45^{o}, 45^{o}

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45^{o}, 90^{o}, 90^{o}

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45^{o}, 135^{o}, 135^{o}

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[\to p + 2 \to q + 3 \to r \to q + 2 \to r + 3 \to p \to r + 2 \to p + 3 \to q] = 54

\to p

\to q

\to r

∣ ∣ ∣ ∣ \begin{matrix} \to p . \to p & \to p . \to q & \to p . \to r \to p . \to q & \to q . \to q & \to q . \to r \to p . \to r & \to r . \to q & \to r . \to r \end{matrix} ∣ ∣ ∣ ∣

\to P \times \to Q = \to R; \to Q \times \to R = \to P

\to R \times \to P = \to Q

\to P, \to Q

\to R

\to P

\to Q

90^{0}

\to P + \to Q + \to R

\to P, \to Q

\to R

P \neq Q \neq R \neq 0

\to R = \to P + \to Q

(\to P \times \to R) = (\to P \times \to Q)

\to q, \to r

\to p = \to q \times \to p + \to r

\to p, \to q, \to r

\to x

\to p \times ((\to x - \to q) \times \to p) + \to q \times ((\to x - \to r) \times \to q) + \to r \times ((\to x - \to p) \times \to r) = 0

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