If overrightarrow{ F }=2 hat{ i }+hat{ j }-ha | vedclass

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(C) Given: $\overrightarrow{ F }=2 \hat{ i }+\hat{ j }-\hat{ k }$ and $\overrightarrow{ r }=3 \hat{ i }+2 \hat{ j }-2 \hat{ k }$.$1$. Scalar Product (

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$10, \sqrt{2}$

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(C) Given: $\overrightarrow{ F }=2 \hat{ i }+\hat{ j }-\hat{ k }$ and $\overrightarrow{ r }=3 \hat{ i }+2 \hat{ j }-2 \hat{ k }$.$1$. Scalar Product (Dot Product):$\overrightarrow{ F } \cdot \overrightarrow{ r } = (2)(3) + (1)(2) + (-1)(-2) = 6 + 2 + 2 = 10$.$2$. Vector Product (Cross Product):$\overrightarrow{ F } 	imes \overrightarrow{ r } = \begin{vmatrix} \hat{ i } & \hat{ j } & \hat{ k } \ 2 & 1 & -1 \ 3 & 2 & -2 \end{vmatrix}$$= \hat{ i }((1)(-2) - (-1)(2)) - \hat{ j }((2)(-2) - (-1)(3)) + \hat{ k }((2)(2) - (1)(3))$$= \hat{ i }(-2 + 2) - \hat{ j }(-4 + 3) + \hat{ k }(4 - 3)$$= 0 \hat{ i } + 1 \hat{ j } + 1 \hat{ k } = \hat{ j } + \hat{ k }$.$3$. Magnitude of Vector Product:$|\overrightarrow{ F } 	imes \overrightarrow{ r }| = \sqrt{0^2 + 1^2 + 1^2} = \sqrt{2}$.Thus,the magnitudes are $10$ and $\sqrt{2}$.

Column $I$	Column $II$
$(A) \|A+B\|$	$(p) \frac{\sqrt{3}}{2} x^2$
$(B) \|A-B\|$	$(q) x$
$(C) A \cdot B$	$(r) \sqrt{3} x$
$(D) \|A \times B\|$	$(s) \frac{x^2}{2}$

If $\overrightarrow{ F }=2 \hat{ i }+\hat{ j }-\hat{ k }$ and $\overrightarrow{ r }=3 \hat{ i }+2 \hat{ j }-2 \hat{ k }$,then the scalar and vector products of $\overrightarrow{ F }$ and $\overrightarrow{ r }$ have the magnitudes respectively as

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