If a = hat{i} + hat{j} + t hat{k} and b = hat | vedclass

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(B) Given vectors are $a = \hat{i} + \hat{j} + t \hat{k}$ and $b = \hat{i} + 2 \hat{j} + 3 \hat{k}$.First,we find the sum $(a+b)$:$a+b = (\hat{i} + \h

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$\pm 2 \sqrt{3}$

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(B) Given vectors are $a = \hat{i} + \hat{j} + t \hat{k}$ and $b = \hat{i} + 2 \hat{j} + 3 \hat{k}$.First,we find the sum $(a+b)$:$a+b = (\hat{i} + \hat{j} + t \hat{k}) + (\hat{i} + 2 \hat{j} + 3 \hat{k}) = 2 \hat{i} + 3 \hat{j} + (t+3) \hat{k}$.Next,we find the difference $(a-b)$:$a-b = (\hat{i} + \hat{j} + t \hat{k}) - (\hat{i} + 2 \hat{j} + 3 \hat{k}) = 0 \hat{i} - \hat{j} + (t-3) \hat{k}$.Since $(a+b)$ and $(a-b)$ are perpendicular,their dot product must be zero:$(a+b) \cdot (a-b) = 0$.$(2 \hat{i} + 3 \hat{j} + (t+3) \hat{k}) \cdot (0 \hat{i} - \hat{j} + (t-3) \hat{k}) = 0$.Calculating the dot product:$(2)(0) + (3)(-1) + (t+3)(t-3) = 0$.$0 - 3 + (t^2 - 9) = 0$.$t^2 - 12 = 0$.$t^2 = 12$.$t = \pm \sqrt{12} = \pm 2 \sqrt{3}$.

If $a = \hat{i} + \hat{j} + t \hat{k}$ and $b = \hat{i} + 2 \hat{j} + 3 \hat{k}$,then the values of $t$ for which $(a+b)$ and $(a-b)$ are perpendicular are:

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