| Column I | Column II |
| a) A.B = 0 | i) θ = 0 |
| b) A.B = +8 | ii) θ = 90o |
| c) A.B = 4 | iii) θ = 180o |
| d) A.B = -8 | iv) θ = 0o |
Answer:
a) matches with ii)
b) matches with i)
c) matches with iv)
d) matches with iii)
As we know that 
![\[\begin{aligned} &\text { (a) } \overrightarrow{ A } \cdot \overrightarrow{ B }=|\overrightarrow{ A } \| \overrightarrow{ B }| \cos \theta=0 \\ &2 \times 4 \cos \theta=0 \\ &\cos \theta=\cos 90^{\circ} \end{aligned}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-f738b4ea6c5f87cb5ad625fb1598cae3_l3.png "Rendered by QuickLaTeX.com")
Thus, Option (a) of column I matches with option (ii) in column II.
(b) Given, 
Implying the formula, 
Solving we get, 
So option (b) of column I matches with option (i) of column II.
![\[\begin{aligned} &\text { (c) } \overrightarrow{ A } \cdot \overrightarrow{ B }=|\overrightarrow{ A } \| \overrightarrow{ B }| \cos \theta=2 \times 4 \cos \theta \\ &=8 \cos \theta \end{aligned}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-29d132beb3a42b27c95a48684392d2a4_l3.png "Rendered by QuickLaTeX.com")
\cos \theta=\frac{4}{8}=\frac{1}{2} \Longrightarrow \cos \theta=\cos 60^{\circ} \Longrightarrow \theta=60^{\circ}
Hence, option (c) of column I matches with option (iv) of column II.
Solving using formula.
![\[\begin{aligned} &\text { (d) } \overrightarrow{ A } \cdot \overrightarrow{ B }=-8 \\ &|\overrightarrow{ A }||\overrightarrow{ B }| \cos \theta=-8 \\ &2 \times 4 \cos \theta=-8 \\ &\cos \theta=-1 \Longrightarrow \cos \theta=\cos 180^{\circ} \Longrightarrow \theta=180^{\circ} \end{aligned}\]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-11632f16e192278e5345cb00c7e64678_l3.png "Rendered by QuickLaTeX.com")
As a result, option (d) in column I corresponds to option (iii) in column II.