The projection of a vector \(\vec{a}\) on vector \(\vec{b}\) is given by
\(\frac{1}{|\vec{b}|} (\vec{a}.\vec{b})\)
\(|\vec{b}|=\sqrt{4^2+3^2}=\sqrt{16+9}\)=5
\(\vec{a}.\vec{b}\)=8(4)-1(3)+0=32-3=29
The projection of vector \(8\hat{i}-\hat{j}+6\hat{k}\) on vector \(4\hat{i}+3\hat{j}\) will be
\(\frac{1}{|\vec{b}|} (\vec{a}.\vec{b})=\frac{1}{5} (29)=\frac{29}{5}\)

Given that: \(\vec{a}=\hat{i}-\hat{j}+3\hat{k}, \,\vec{b}=5\hat{i}-2\hat{j}+\hat{k} \,and \,\vec{c}=\hat{i}-\hat{j}\)
Also given, \(\vec{a}+?\vec{b}\) is perpendicular to \(\vec{c}\)
Therefore, \((\vec{a}+?\vec{b}).\vec{c}=0\)
i.e. \((\hat{i}-\hat{j}+3\hat{k}+?(5\hat{i}-2\hat{j}+\hat{k})).(\hat{i}-\hat{j})\)=0
\(((1+5?) \,\hat{i}-(1+2?) \,\hat{j}+(?+3) \,\hat{k}).(\hat{i}-\hat{j})\)=0
1+5?+1+2?=0
?=-\(\frac{7}{2}\).

6 Newton is a vector quantity as it is a force. Force is a vector quantity which has both magnitude and direction.

As both vectors are equal hence, we can equate ir constants and get value of x, y and z. Now we equate coefficients of \(\hat{i}\), \(\hat{j}\), \(\hat{k}\) of both equations and get values x=2, y=2, z=1.

The vectors which start from same initial point are called coinitial vectors. In given figure, vectors \(\vec{a}\) and \(\vec{c}\) start from same point but are of different magnitude. Therefore, vectors \(\vec{a}\) and \(\vec{c}\) ae coinitial but not equal.

Given that, \(\vec{a}\)=2\(\hat{i}\)+7\(\hat{j}\) and \(\vec{b}\)=\(\hat{i}\)-9\(\hat{j}\)
The sum of two vectors will be
\(\vec{a}+\vec{b}\)=(2\(\hat{i}\)+7\(\hat{j}\))+(\(\hat{i}\)-9\(\hat{j}\))
=(2+1) \(\hat{i}\)+(7-9)\(\hat{j}\)
=3\(\hat{i}\)-2\(\hat{j}\)
The unit vector in direction of sum of vectors is
\(\frac{1}{|\vec{a}+\vec{b}|} (\vec{a}+\vec{b})=\frac{3\hat{i}-2\hat{j}}{\sqrt{3^2+(-2)^2}}=\frac{3\hat{i}-2\hat{j}}{\sqrt{13}}=\frac{3}{1\sqrt{3}} \hat{i}-\frac{2}{\sqrt{13}}\hat{j}\)

Given that, \(\vec{a}=-\hat{j}+\hat{k}\) and \(\vec{b}=-\hat{i}-\hat{j}-\hat{k}\)
Calculating vector product, we get
\(\vec{a}×\vec{b}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\0&-1&1\\-1&-1&-1\end{vmatrix}\)
=\(\hat{i}(1-(-1))-\hat{j}(0-(-1))+\hat{k}(0-1)\)
=\(2\hat{i}-\hat{j}-\hat{k}\)

\(|\vec{a}+\vec{b}|^2=(\vec{a}+\vec{b}).(\vec{a}+\vec{b})\)
=\(\vec{a}.\vec{a}+\vec{a}.\vec{b}+\vec{b}.\vec{a}+\vec{b}.\vec{b}\)
=\(|\vec{a}|^2+2(\vec{a}.\vec{b})+|\vec{b}|^2\)
=(3)²+2(6)+(4)²
=9+12+16=37
&re4;\(|\vec{a}+\vec{b}|=\sqrt{37}\)

As we know that magnitude of vector is calculated by formula \(\sqrt{x^2+ y^2}\).
Therefore, |\(\vec{a}\)| = \(\sqrt{12} + 22 = \sqrt{5}\) and \(|\vec{b}| = \sqrt{22} + 12 = \sqrt{5}\), y are equal.

The given statement is false. Collinear vectors are those vectors that are parallel same line irrespective of direction and magnitude. The vectors which have same initial point are called coinitial vectors.