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B
|→b||→a|+|→b|→a+|→a||→a|+|→b|→b
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C
|→a||→a|+2|→b|→a+|→b||→a|+2|→b|→b
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D
|→b|2|→a|+|→b|→a+∣∣→a∣∣2|→a|+∣∣∣→b∣∣∣→b
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Solution
The correct options are B|→b||→a|+|→b|→a+|→a||→a|+|→b|→b C|→b|2|→a|+|→b|→a+∣∣→a∣∣2|→a|+∣∣∣→b∣∣∣→b Since →a and →b are equally inclined to →c,→c must be of the form t⎛⎜⎝→a|→a|+→b|→b|⎞⎟⎠ Now |→b||→a|+|→b|→a+|→a||→a|+|→b|→b =|→a||→b||→a|+→b⎛⎜⎝→a|→a|+→b|→a|+|→b|⎞⎟⎠ Also, |→b|2|→a|+|→b|→a+|→a|2|→a|+|→b|→b =|→a|+|→b|2|→a|+→b⎛⎜⎝→a|→a|+→b|→b|⎞⎟⎠ Other two vectors cannot be written in the form t⎛⎜⎝→a|→a|+→b|→b|⎞⎟⎠