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3 years ago

Hassam plotted Point M in the coordinate plane below

Mathematics

2 answers:

Talja [164]3 years ago

5 0

Answer:

the second option

Step-by-step explanation:

pls mark brainliest

ki77a [65]3 years ago

4 0

Answer:

The answer to your question is the second option, both coordinates should be negative.

Step-by-step explanation:

The point Hassam plotted was (-0.5, -0.25). If he wanted to plot point (0.5, 0.25):

-First, he should have counted two squares to the right

- then one square up.

Because both coordinates are positive this point must be in the first quadrangle not in the third.

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We have been given two vectors $\vec{a}$ and $\vec{b}$ , we are to find out the scalar and vector projection of $\vec{b}$ onto $\vec{a}$

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$\begin{aligned}&=\frac{(\hat{i}+\hat{j}+\hat{k}) \cdot(\hat{i}-\hat{j}+\hat{k})}{\sqrt{1^2+1^1+1^2}} \\&=\frac{1^2-1^2+1^2}{\sqrt{3}}=\frac{1}{\sqrt{3}}\end{aligned}$

The Vector projection of $\vec{b}$ onto $\vec{a}$ means the resolved component of $\vec{b}$ in the direction of $\vec{a}$ and is given by

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$\vec{a}=\frac{\vec{b} \cdot \vec{a}}{|\vec{a}|^2} \cdot(\hat{i}+\hat{j}+\hat{k})$

$\begin{aligned}&=\frac{(\hat{i}+\hat{j}+\hat{k}) \cdot(\hat{i}-\hat{j}+\hat{k})}{\left(\sqrt{1^2+1^1+1^2}\right)^2} \cdot(\hat{i}+\hat{j}+\hat{k}) \\&=\frac{1^2-1^2+1^2}{3} \cdot(\hat{i}+\hat{j}+\hat{k})=\frac{1}{3}(\hat{i}+\hat{j}+\hat{k})\end{aligned}$

To learn more about scalar and vector projection visit:brainly.com/question/21925479

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1 year ago

Hassam plotted Point M in the coordinate plane below​

Hassam plotted Point M in the coordinate plane below