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We know that cross product of two vectors gives us a vector which is perpendicular to both the vectors. And keeping in mind that is a Unit vector we get the equation –

\(\cfrac{\vec a\times\vec b}{|\vec a\times\vec b|}=\vec c\)→(Vector divided its magnitude gives unit vector)

\(\cfrac{\vec a\times\vec b}{|\vec a\times\vec b|}=-\vec c\) \(\therefore-\vec c\) is perpendicular to \(\vec a\) and  \(\vec b\).

Alternative Solution – Since \(\vec c\) is perpendicular to \(\vec a\) and  \(\vec b\), any unit vector parallel/anti-parallel to  \(\vec c\) will be perpendicular to \(\vec a\) and  \(\vec b\).

Correct option is B. an ellipse

Let  \(\vec a=\hat i-2x\hat j+2y\hat k\) and  \(\vec b=\hat i+2xj-3y\hat k\)

Given that  \(\vec a\) and  \(\vec b\) are perpendicular.

So,  \(\vec a-\vec b\) = 0

⇒ 1 - 4x² - 6y² = 0

⇒ 4x²+ 6y² =1

Here, vectors are in 3-Dimensions

∴  above equation represents an ellipse .i.e. locus of (x, y) is an ellipse.

 \(\vec a.\vec b=|a||b|cos\theta=6\)

Given,

|a|=3, |b|=4

6 = 3 x 4 cos θ

6 = 12 cos θ

cos θ = \(\cfrac12\)

We know,

\(\vec a.\vec b=|a||b|cos\theta\) where θ is the angle between \(\vec a\) and  \(\vec b\).

Given, |a |= 2 |b| = √3

\(\vec a.\vec b=2.\sqrt3cos\theta\)

So, cos θ = \(\cfrac12\)

θ = 60°

Correct option is B. \(\vec a.\vec b=0\) 

The diagonals of a rhombus are always perpendicular

It means  \(\vec a\) is perpendicular to \(\vec b\)

Q = 90°

Cos Q = 0

\(\vec a.\vec b=0\)

Let,

 \(\vec{a} =( \vec{8i} + \vec{j})\)

\(\vec{b} = (\vec{i} + \vec{2j} - \vec{2k})\)

\(\vec{b}\) = \(\sqrt{1^2 + 2^2+ 2^2}\) = \(\sqrt{1+ 4+ 4}\) = \(\sqrt {9}\) = 3

 \(\vec{b}\) = \(\frac{\vec{b}}{\vec{|b|}}\) = \(\frac{\vec{i}+\vec{2j}-\vec{2k}}{3}\) 

∴ The projection of \((\vec{8i}+\vec{j})\) on \((\vec{i}+{2j}-\vec{2k})\) is   \((\vec{8i}+\vec{j})\)   \(\frac{\vec{i}+\vec{2j}-\vec{2k}}{3}\) 

= \(\frac{8+2+0}{3}\) = \(\frac{10}{3}\)

\((\vec{a}+\vec{b}).(\vec{a}-\vec{b})\) = \(|\vec{a}|^2+|\vec{b}|^2\) 

⇒  \(|\vec{a}|^2-|\vec{b}|^2\) =   \(|\vec{a}|^2+|\vec{b}|^2\)

⇒    \(|\vec{b}|\) = 0

Which is not possible hence

\((\vec{a})⊥(\vec{b})\)

As we know, for vectors  \(\vec a\) and  \(\vec b\) unit vectors perpendicular to them is give by \(\pm\cfrac{\vec a\times\vec b}{|\vec a\times\vec b|}\) 

Unit vector can be ⊥ either in positive or negative direction.

Hence, the number of vectors of unit length perpendicular to both the vectors  \(\vec a=2\hat i+\hat j+2\hat k\) and  \(\vec b = \hat j+\hat k\)  is 2.

Given question gives us two same vectors so the angle 0°.

 is In case, it asks write the angle between the vectors \(\vec a\times\vec b\) and \(\vec b\times \vec a-\)

The angle between the vectors will be 180 ° as they are equal in magnitude and opposite in direction.

Given  \(\vec a\times\vec b=\vec c,\)  \(\vec b\times\vec c=\vec a\) and  \(\vec c\times\vec a=\vec d\).

Considering the first equation, \(\vec c\) is the cross product of the vectors \(\vec a\) and  \(\vec b\).

By the definition of the cross product of two vectors, we have  \(\vec c\) perpendicular to both \(\vec a\) and  \(\vec b.\).

Similarly, considering the second equation, we have  \(\vec a\) perpendicular to both \(\vec b\) and  \(\vec c\).

Once again, considering the third equation, we have  \(\vec b\) perpendicular to both \(\vec c\) and  \(\vec a\).

From the above three statements, we can observe that the vectors \(\vec a,\,\vec b\) and \(\vec c\) are mutually perpendicular.

It is also said that  \(\vec a,\,\vec b\) and  \(\vec c\) are three unit vectors.

Thus, \(\vec a,\,\vec b,\,\vec c\) form an orthonormal right handed triad of unit vectors.

x - axis = \(\hat i\)

y - axis = \(\hat j\)

z - axis = \(\hat k\)

Proj \(\vec b\)^{\(\vec a\)} =  \(\cfrac{\vec a.\vec b}{|\vec b|^2}\vec b\)

Projection along x - axis = \(\cfrac31\hat i\)

= 3 \(\hat i\)

Projection along y - axis = \(\cfrac{-4}1\hat j\)

= -4 \(\hat j\)

Projection along z - axis  = \(\cfrac{12}1\hat k\)

 = 12 \(\hat k\)

\(\vec a.(\vec b\times \vec a)=0\)

We know,

\((\vec b\times \vec a)\) is perpendicular to both  \(\vec a\) and  \(\vec b\).

So, \(\vec a.(\vec b\times \vec a)=0\)

[\(\because\vec a\) and \((\vec b\times\vec a)\) are perpendicular to each other]

\((\vec a\times\vec b).\vec b=0\)

We know, \((\vec a\times\vec b)\) is perpendicular to both  \(\vec a\) and  \(\vec b\).

So, \((\vec a\times\vec b).\vec b=0\)

[\(\because\vec b\) and \((\vec a\times\vec b)\) are perpendicular to each other]

Let,

\(\vec{a}= (\vec{i}+\vec{j}+\vec{k})\)

\(\vec{b}=(\vec{j})\)

\(\vec{|b|}\) = \(\sqrt{0^2+1^2+0^2}\) = \(\sqrt{1}\) = 1

 \(\vec{b}=\frac{\vec{b}}{|\vec{b}|}\) = \(\frac{\vec{j}}{1}\)

∴ The projection of \((\vec{i}+\vec{j}+\vec{k})\) on \((\vec{j})\) is   \((\vec{i}+\vec{j}+\vec{k})\).\((\vec{j})\) = 1

it is given that  \(\vec a.\vec a=0\) and  \(\vec a.\vec b=0\)

From this, we can say that  \(|\vec a|^2=0\)

So,  \(\vec a\)  is a zero vector \(\vec a.\vec b=0\) we can say that  \(\vec b\) can be any vector perpendicular to zero vector  \(\vec a.\) 

Volume of the parallelepiped with its edges represented by the vectors

\(\vec a,\vec b,\vec c\) is \([\vec a\,\vec b\,\vec c]\) = \(\vec a.(\vec b\times\vec c)\) = \(\begin{vmatrix} 1 & 1 & 0 \\ 1 & 2 & 0 \\ 1 & 1 & \pi \end{vmatrix}\) = π

Three vectors are coplanar if (if and only if)  \(\vec a.(\vec b\times\vec c)=0\)

Hence we have value of the matrix \(\begin{vmatrix}1 & 2 & 1 \\a & 1 & 2 \\1 & 2 & a\end{vmatrix}\) = 0

We have 2a² - 3a + 1 = 0

2a² - 2a - a + 1 = 0

Solving this quadratic equation we get

a = 1, a = \(\cfrac12\)