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Suppose that height=17 Does this mean that we must have either d ≥ d1 or d ≥ d­2 ­? If not, explain why.

Shane Macguire , 10 Years ago
Grade upto college level
anser 1 Answers
Deepak Patra
It is not necessary that the if\overrightarrow{d} = \overrightarrow{d_{1}} + \overrightarrow{d _{2}} , then the magnitude of vector\overrightarrow{d} will be either greater than or equal to magnitude of vector\overrightarrow{d_{1}} or vector\overrightarrow{d_{2}} respectively.
Consider a two dimensional vector\overrightarrow{d_{1}} with componenta_{x} along the unit vector in x-direction,a_{y} along the unit vector in y-direction respectively.
Mathematically, the vector\overrightarrow{d_{1}} can be represented as:
\overrightarrow{d_{1}} = a_{x}\widehat{i}+ a_{y}\widehat{j}
The magnitude of the vector\overrightarrow{d_{1}} is given as:
|\overrightarrow{d_{1}}| = \sqrt{a_{x}^{2}+ a_{y}^{2}}
Consider another vector\overrightarrow{d_{1}} with componentb_{x} along the unit vector in x-direction, b_{y}along the unit vector in y-direction respectively.
Mathematically, the vector\overrightarrow{d_{2}} can be represented as:
\overrightarrow{d_{2}} = b_{x}\widehat{i}+b_{y}\widehat{j}
The magnitude of the vector\overrightarrow{d_{2}} is given as:
|\overrightarrow{d_{2}}| = \sqrt{b_{x}^{2} + b_{y}^{2}}
The sum of vector\overrightarrow{d_{1}} and vector\overrightarrow{d_{2}} is given as:


233-74_12.PNG

One can see from above that the magnitude of vector\overrightarrow{d} depends on the value of (a_{x}+b_{x}) and (a_{y}+b_{y}). Therefore, in a situation where has opposite sign to that withb_{x} anda_{y} has the opposite sign to that withb_{y} , the magnitude of vector\overrightarrow{d} will be smaller than the magnitude of either vector\overrightarrow{d_{1}} or vector \overrightarrow{d_{2}}.
For example, let us assume that vector\overrightarrow{d_{1}} is given as:
\overrightarrow{d_{1}} = 5\widehat{i} + 5\widehat{j}
And the vector\overrightarrow{d_{2}} is given as:
\overrightarrow{d_{2}} = -2\widehat{i}-4\widehat{j}
The magnitude of vector \overrightarrow{d_{1}} is:
|\overrightarrow{d_{1}}| = \sqrt{(5)^{2} + (5)^{2}}
= \sqrt{50}
The magnitude of vector\overrightarrow{d_{2}} is:
|\overrightarrow{d_{2}}| = \sqrt{(-2)^{2}+(-4)^{2}}
= \sqrt{20}
The magnitude of vector \overrightarrow{d} is:
|\overrightarrow{d}| = \sqrt{(5-2)^{2} + (5-4)^{2}}
= \sqrt{(3)^{2} + (1)^{2}}
= \sqrt{10}
Therefore the magnitude of vector\overrightarrow{d} is smaller than the magnitude of both vector \overrightarrow{d^{1}}and vector\overrightarrow{d^{2}} . It is important to note that in this assumption, the vector\overrightarrow{d^{2}} is such, that the value(a_{x}+ b_{x}) of and(a_{y}+ b_{y}) is reduced significantly and the magnitude of vector \overrightarrow{d}became smaller than the magnitude of both the vectors\overrightarrow{d_{1}} and\overrightarrow{d_{2}} . However if we take our vector\overrightarrow{d_{2}} as say --\widehat{i}-\widehat{j}, the magnitude of vector\overrightarrow{d} will be greater than the magnitude of vector\overrightarrow{d_{2}} .



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