If PQ + QR = (2lambda^2 - 5)RP,then lambda is | vedclass

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(B) Given the vector equation: $PQ + QR = (2\lambda^2 - 5)RP$ By the triangle law of vector addition,we know that $PQ + QR = PR$. Substituting this in

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$\pm \sqrt{2}$

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(B) Given the vector equation: $PQ + QR = (2\lambda^2 - 5)RP$ By the triangle law of vector addition,we know that $PQ + QR = PR$. Substituting this into the given equation,we get: $PR = (2\lambda^2 - 5)RP$ Since $PR = -RP$,we can rewrite the equation as: $-RP = (2\lambda^2 - 5)RP$ Dividing both sides by $RP$ (assuming $RP 
eq 0$),we get: $2\lambda^2 - 5 = -1$ $2\lambda^2 = 4$ $\lambda^2 = 2$ $\lambda = \pm \sqrt{2}$

If $PQ + QR = (2\lambda^2 - 5)RP$,then $\lambda$ is equal to

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