Given $a+b+c+d=0,$ which of the following statements eare correct:
$(a)\;a, b,$ c, and $d$ must each be a null vector,
$(b)$ The magnitude of $(a+c)$ equals the magnitude of $(b+d)$
$(c)$ The magnitude of a can never be greater than the sum of the magnitudes of $b , c ,$ and $d$
$(d)$ $b + c$ must lie in the plane of $a$ and $d$ if $a$ and $d$ are not collinear, and in the line of a and $d ,$ if they are collinear ?
$(a)$ Incorrect : In order to make $a+b+c+d=0,$ 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.
$(b)$ 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:
$| a |=| b + c + d |$
$| a | \leq| a |+| b |+| c | \ldots \ldots(i)$
Equation $(i)$ 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 $a+b+c+d=0$
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.
When vector $\overrightarrow{ A }=2 \hat{ i }+3 \hat{ j }+2 \hat{ k }$ is subtracted from vector $\vec{B}$, it gives a vector equal to $2 \hat{j}$. Then the magnitude of vector $\vec{B}$ will be:
$ABC$ is an equilateral triangle. Length of each side is $a$ and centroid is point $O$. Find If $|\overrightarrow{A B}+\overrightarrow{B C}+\overrightarrow{A C}|=n a$ then $n =$ ?
Prove the associative law of vector addition.
Mark the correct statement :-