If overrightarrow{A} = hat{i} + 2hat{j} + 3ha | vedclass

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(C) Let $\overrightarrow{V} = \overrightarrow{A} + t\overrightarrow{B}$.$\overrightarrow{V} = (\hat{i} + 2\hat{j} + 3\hat{k}) + t(-\hat{i} + 2\hat{j}

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(C) Let $\overrightarrow{V} = \overrightarrow{A} + t\overrightarrow{B}$.$\overrightarrow{V} = (\hat{i} + 2\hat{j} + 3\hat{k}) + t(-\hat{i} + 2\hat{j} + \hat{k})$$\overrightarrow{V} = (1 - t)\hat{i} + (2 + 2t)\hat{j} + (3 + t)\hat{k}$.Given that $\overrightarrow{V}$ is perpendicular to $\overrightarrow{D} = 3\hat{i} + 4\hat{j}$,their dot product must be zero.$\overrightarrow{V} \cdot \overrightarrow{D} = 0$$3(1 - t) + 4(2 + 2t) + 0(3 + t) = 0$$3 - 3t + 8 + 8t = 0$$11 + 5t = 0$$5t = -11$$t = -2.2$Note: Re-evaluating the dot product with the provided vector $3\hat{i} + \hat{j}$ from the original text:$3(1 - t) + 1(2 + 2t) = 0$$3 - 3t + 2 + 2t = 0$$5 - t = 0 \Rightarrow t = 5$.

If $\overrightarrow{A} = \hat{i} + 2\hat{j} + 3\hat{k}$,$\overrightarrow{B} = -\hat{i} + 2\hat{j} + \hat{k}$ and $\overrightarrow{C} = 3\hat{i} + \hat{j}$,then the value of $t$ such that $\overrightarrow{A} + t\overrightarrow{B}$ is at a right angle to vector $3\hat{i} + 4\hat{j}$ is

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