Which of these can not form a perpendicular bisector?v PLZ HALLLLPPPP

Mathematics

2 answers:

polet [3.4K]3 years ago

4 0

A circle obviously. A perpendicular bisector requires right angles which a circle can physically never have.

jasenka [17]3 years ago

3 0

Answer: Second option is correct.

Step-by-step explanation:

Since we know that to form perpendicular bisector .

We need atleast two lines which form right angle.

So, right triangle can form perpendicular bisector with base and height.

Parallel lines can also form perpendicular bisector if a line is drawn from mid point of one line to mid point of another line.

A line segment can form perpendicular bisector.

But circle can't as it is curved shape it needs line to make perpendicular bisector.

Hence, Second option is correct.

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Find a nonzero vector orthogonal to the plane through the points: ????=(0,0,1), ????=(−2,3,4), ????=(−2,2,0).

ser-zykov [4K]

Answer:

The nonzero vector orthogonal to the plane is <-9,-8,2>.

Step-by-step explanation:

Consider the given points are P=(0,0,1), Q=(−2,3,4), R=(−2,2,0).

$\overrightarrow {PQ}==$

$\overrightarrow {PR}==$

The nonzero vector orthogonal to the plane through the points P,Q, and R is

$\overrightarrow n=\overrightarrow {PQ}\times \overrightarrow {PR}$

$\overrightarrow n=\det \begin{pmatrix}i&j&k\\ \:\:\:\:\:-2&3&3\\ \:\:\:\:\:-2&2&-1\end{pmatrix}$

Expand along row 1.

$\overrightarrow n=i\det \begin{pmatrix}3&3\\ 2&-1\end{pmatrix}-j\det \begin{pmatrix}-2&3\\ -2&-1\end{pmatrix}+k\det \begin{pmatrix}-2&3\\ -2&2\end{pmatrix}$