If overrightarrow{AB} = (2hat{i} + 3hat{j} - | vedclass

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(C) Let the coordinates of point $B$ be $(x, y, z)$.Given the position vector of point $A$ is $\vec{a} = (1\hat{i} + 2\hat{j} - 1\hat{k})$.The vector

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$(3, 5, -2)$

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(C) Let the coordinates of point $B$ be $(x, y, z)$.Given the position vector of point $A$ is $\vec{a} = (1\hat{i} + 2\hat{j} - 1\hat{k})$.The vector $\overrightarrow{AB}$ is given by $\vec{b} - \vec{a}$.So,$\overrightarrow{AB} = (x - 1)\hat{i} + (y - 2)\hat{j} + (z - (-1))\hat{k} = (x - 1)\hat{i} + (y - 2)\hat{j} + (z + 1)\hat{k}$.We are given $\overrightarrow{AB} = 2\hat{i} + 3\hat{j} - 1\hat{k}$.Comparing the components,we get:$x - 1 = 2 \Rightarrow x = 3$$y - 2 = 3 \Rightarrow y = 5$$z + 1 = -1 \Rightarrow z = -2$Therefore,the coordinates of $B$ are $(3, 5, -2)$.

If $\overrightarrow{AB} = (2\hat{i} + 3\hat{j} - \hat{k})$ and $A(1, 2, -1)$ is the given point,then the coordinates of $B$ are

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