If |a| = 3 and |b| = 4,for what value of lamb | vedclass

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(B) Two vectors are perpendicular if their dot product is $0$.Given that $(a + \lambda b)$ is perpendicular to $(a - \lambda b)$,we have:$(a + \lambda

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$3/4$

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(B) Two vectors are perpendicular if their dot product is $0$.Given that $(a + \lambda b)$ is perpendicular to $(a - \lambda b)$,we have:$(a + \lambda b) \cdot (a - \lambda b) = 0$Using the property of the dot product $(u + v) \cdot (u - v) = |u|^2 - |v|^2$,we get:$|a|^2 - |\lambda b|^2 = 0$$|a|^2 - \lambda^2 |b|^2 = 0$Substitute the given values $|a| = 3$ and $|b| = 4$:$3^2 - \lambda^2 (4^2) = 0$$9 - 16\lambda^2 = 0$$16\lambda^2 = 9$$\lambda^2 = \frac{9}{16}$$\lambda = \pm \frac{3}{4}$

If $|a| = 3$ and $|b| = 4$,for what value of $\lambda$ is the vector $(a + \lambda b)$ perpendicular to $(a - \lambda b)$?

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