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(C) We know that the dot product of two perpendicular vectors is zero. Since $(\vec{a}+t \vec{b})$ and $(\vec{a}-t \vec{b})$ are perpendicular,their d

Vedclass · Accepted Answer

$t=\frac{2}{3}$

Vedclass · Answer

(C) We know that the dot product of two perpendicular vectors is zero. Since $(\vec{a}+t \vec{b})$ and $(\vec{a}-t \vec{b})$ are perpendicular,their dot product is zero: $(\vec{a}+t \vec{b}) \cdot (\vec{a}-t \vec{b}) = 0$ Expanding the dot product using the distributive property: $|\vec{a}|^2 - t(\vec{a} \cdot \vec{b}) + t(\vec{b} \cdot \vec{a}) - t^2|\vec{b}|^2 = 0$ Since the dot product is commutative,$\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$,so the middle terms cancel out: $|\vec{a}|^2 - t^2|\vec{b}|^2 = 0$ Substituting the given values $|\vec{a}|=2$ and $|\vec{b}|=3$: $2^2 - t^2(3^2) = 0$ $4 - 9t^2 = 0$ $9t^2 = 4$ $t^2 = \frac{4}{9}$ $t = \pm \frac{2}{3}$ Since $t$ is given as a positive scalar,we have $t = \frac{2}{3}$.

If $\vec{a}$ and $\vec{b}$ are two vectors such that $|\vec{a}|=2$,$|\vec{b}|=3$ and $\vec{a}+t \vec{b}$ and $\vec{a}-t \vec{b}$ are perpendicular,where $t$ is a positive scalar,then

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