The given statement is true. The vector whose magnitude is unity is called a unit vector. The unit vector in given direction of given vector \(\vec{b}\) is denoted by \(\hat{b}\).

Consider \((\vec{a}-\vec{b})×(\vec{a}+\vec{b})\)
=\((\vec{a}-\vec{b})×\vec{a}+(\vec{a}+\vec{b})×\vec{b}\)
=\(\vec{a}×\vec{a}-\vec{b}×\vec{a}+\vec{a}×\vec{b}-\vec{b}×\vec{b}\)
We know that, \(\vec{a}×\vec{a}=0,\vec{b}×\vec{b}=0 \,and \,\vec{a}×\vec{b}=-\vec{b}×\vec{a}\)
&re4; \(\vec{a}×\vec{a}-\vec{b}×\vec{a}+\vec{a}×\vec{b}-\vec{b}×\vec{b}=0+2(\vec{a}×\vec{b})+0\)
Hence, \((\vec{a}-\vec{b})×(\vec{a}+\vec{b})=2(\vec{a}×\vec{b})\)

Given that, \(\vec{a}\)=4\(\hat{i}\)-4\(\hat{j}\), \(\vec{b}\)=-3\(\hat{i}\)+2k, \(\vec{c}\)=7\(\hat{j}\)-8\(\hat{k}\)
To find: \(\vec{a}+\vec{b}\)+\(\vec{c}\)
&re4;\(\vec{a}+\vec{b}\)+\(\vec{c}\)=(4\(\hat{i}\)-4\(\hat{j}\)) +(-3\(\hat{i}\)+2k) +(7\(\hat{j}\)-8\(\hat{k}\))
=(4-3) \(\hat{i}\)+(-4+7) \(\hat{j}\)+(2-8)\(\hat{k}\)
=\(\hat{i}\)+3\(\hat{j}\)-6\(\hat{k}\)

The projection of vector \(\vec{b}\) on vector \(\vec{b}\) is given by \(\frac{1}{|\vec{a}|} (\vec{a}.\vec{b})\)
\(|\vec{a}|=\sqrt{(1)^2+(-1)^2+(-\sqrt{2})^2}=\sqrt{1+1+2}=\sqrt{4}\)=2
Also, \(\vec{a}.\vec{b}=2(1)+2\sqrt{2} \,(-1)-2(-\sqrt{2})=2-2\sqrt{2}+2\sqrt{2}\)=2
Therefore, projection of vector \(\hat{i}-\hat{j}-\sqrt{2} \,\hat{k}\) on vector \(\vec{b}=2\hat{i}+2\sqrt{2}\hat{j}-2\hat{k}\) is
\(\frac{1}{|\vec{a}|} (\vec{a}.\vec{b})=\frac{1}{2}\) (2)=1.

It satisfies distribution property over addition, hence in k \(\vec{a}\) + m\(\vec{a}\) we can take vector \(\vec{a}\)
common and answer come out to be (k+m)\(\vec{a}\). Basically it?s a simplification method by which vectors can be easily solved and furr properties can be applied to m.

Speed is a vector quantity as it has both magnitude and direction. Time, weight, volume have only magnitude and no direction. Hence, y all are scalar quantity.

Consider \(\vec{a}\) is position vector of a point M(x,y,z) and ?, ?, ? are angles, made by vector \(\vec{a}\) with positive directions of x, y and z respectively. The cosines of angles, cos??, cos??, cos?? are direction cosines of vector \(\vec{a}\) denoted by l, m, n, n
cos²??+cos²??+cos²??=1 i.e.l²+m²+n²=1.

Given that, \(\vec{a}=2\hat{i}+4\hat{j}\) and \(\vec{b}=3\hat{i}-\hat{j}+2\hat{k}\)
Calculating vector product, we get
\(\vec{a}×\vec{b}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\2&4&-5\\3&-1&2\end{vmatrix}\)
=\(\hat{i}(8-5)-\hat{j}(4-(-15))+\hat{k}(-2-12)\)
=\(3\hat{i}-19\hat{j}-14\hat{k}\)

Given that, \(\vec{a}\)=8\(\hat{i}\)+5\(\hat{j}\) and \(\vec{b}\)=-2\(\hat{i}\)+6\(\hat{j}\)
&re4;The sum of vectors will be
\(\vec{a}+\vec{b}\)=(8\(\hat{i}\)+5\(\hat{j}\))+(-2\(\hat{i}\)+6\(\hat{j}\))
=(8-2) \(\hat{i}\)+(5+6)\(\hat{j}\)
=6\(\hat{i}\)+11\(\hat{j}\)

Two vectors are said to be equal, if y have same magnitude. In given figure, vectors \(\vec{a}\) and \(\vec{b}\) have same magnitude. Hence, y are equal.