Given $\overline { a } + \overline { b } + \vec { c } + \overline { d } = 0$ , which of the following statements is/are not a correct statement?
- $\vec { a } , \vec { b } , \vec { c }$ and $\vec { d }$ must be a null vector.
- The magnitude of $( \vec { a } + \vec { c } )$ equals the magnitude of $a( \vec { b } + \vec { d } )$
- The magnitude of $\vec { a }$ can never be greater than the sum of the magnitudes of $\vec { b } , \vec { c }$ and $\vec { d }$
- $\vec{b}$+$\vec{c}$ must He in the plane of $\vec{a}$ and $\vec{d}$ if $\vec{a}$ and $\vec{d}$ are not collinear and in the line of $\vec{a}$ and $\vec{d}$, if they are collinear.
In order to make vectors a + b + c + d = 0, it is not necessary to have all the four given vectors to be null vectors. There are many other combinations which can give the sum zero.
Simple example:- a=î,b=2î,c=–3î,d=0
(b) Correct
a + b + c + d = 0
a + c = – (b + d)
Taking modulus on both the sides, we get:
| a + c | = | –(b + d)| = | b + d |
Hence, the magnitude of (a + c) is the same as the magnitude of (b + d).
(c) Correct
a + b + c + d = 0
a = – (b + c + d)
Taking modulus both sides, we get:
| a | = | b + c + d |
| a | ≤ | a | + | b | + | c | …. (#)
Equation (#) shows that the magnitude of a is equal to or less than the sum of the magnitudes of b, c, and d.
Hence, the magnitude of vector a can never be greater than the sum of the magnitudes of b, c, and d.
(d) Correct
For a + b + c + d = 0
a + (b + c) + d = 0
The resultant sum of the three vectors a, (b + c), and d can be zero only if (b + c) lie in a plane containing a and d, assuming that these three vectors are represented by the three sides of a triangle.
If a and d are collinear, then it implies that the vector (b + c) is in the line of a and d. This implication holds only then the vector sum of all the vectors will be zero.
This is the explanation of correct solution.