Prove that \(\left[ \overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } ,\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } ,\overset { \rightarrow }{ c } \right] \)=\(\left[ \overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] \)

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Prove by vector method, that in a right angled triangle the square of the hypotenuse is equal to the sum of the square of the other two sides.

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Cartesian equation

If \(\overset { \rightarrow }{ a } +\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } =0\) then show that \(\overset { \rightarrow }{ a } \times \overset { \rightarrow }{ b } =\overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } =\overset { \rightarrow }{ c } \times \overset { \rightarrow }{ a } \)

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lies in the plane containing \(\overset { \rightarrow }{ b } \) and \(\overset { \rightarrow }{ c } \)

Show that the four points whose position vectors are \(6\overset { \wedge }{ i } -7\overset { \wedge }{ j } ,16\overset { \wedge }{ i } -29\overset { \wedge }{ j } -4\overset { \wedge }{ k } ,3\overset { \wedge }{ i } -6\overset { \wedge }{ j } \) are co-planar

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Show that the lines \(\frac { x-1 }{ 3 } =\frac { y+1 }{ 2 } =\frac { z-1 }{ 5 } \) and \(\frac { x+2 }{ 4 } =\frac { y-1 }{ 3 } =\frac { z+1 }{ -2 } \) do not intersect

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a, b, c

Find the angle between the vectors \(3 \hat{i}-2 \hat{j}-6 \hat{k}\) and \(4 \hat{i}-\hat{j}+8 \hat{k}\)

Find the angle between the vectors \(\vec a\) and \(\vec b\) where \(\vec{a}=\hat{i}-\hat{j} \text { and } \vec{b}=\vec{j}-\vec{k}\)

For any two vector \(\vec a\) and \(\vec b\) prove that \(|\vec{a}+\vec{b}|^{2}+|\vec{a}-\vec{b}|^{2}=2\left(|\vec{a}|^{2}+|\vec{b}|^{2}\right)\)

Show that the vector \(2 \hat{i}-\hat{j}+\hat{k}, \hat{i}-3 \hat{j}-5 \hat{k},-3 \hat{i}+4 \hat{j}+4 \hat{k}\) form the sides of a right angled triangle.

If \(|\vec{a}|=13,|\vec{b}|=5\) and \(\vec{a} \cdot \vec{b}=60\) then find \(|\vec{a} \times \vec{b}|\)

If \(\vec{a}, \vec{b}\) are any two vectors, then \(|a \times b|^{2}+(a \cdot b)^{2}=\) \(|\vec{a}|^{2}|\vec{b}|^{2}\)

Show that the points (1, 3, 1), (1, 1, -1), (-1, 1, 1) and (2, 2,-1) are lying on the same plane. (Hint : It is enough to prove any three vectors formed by these four points are coplanar).

If \(\vec{x} \cdot \vec{a}=0, \vec{x} \cdot \vec{b}=0, \vec{x} \cdot \vec{c}=0 \text { and } \vec{x} \neq \overrightarrow{0}\)then show that \(\vec{a}, \vec{b}, \vec{c}\) are coplanar.

Let  \(\vec{a}, \vec{b}, \vec{c} \text { and } \vec{d}\) be any four vectors then
\((i)\ (\vec{a} \times \vec{b}) \times(\vec{c} \times \vec{d})=[\vec{a}, \vec{b}, \vec{d}] \vec{c}-[\vec{a}, \vec{b}, \vec{c}] \vec{d} \)
\((ii)\ (\vec{a} \times \vec{b}) \times(\vec{c} \times \vec{d})=[\vec{a}, \vec{c}, \vec{d}] \vec{b}-[\vec{b}, \vec{c}, \vec{d}] \vec{a}\)

Prove that \(\vec{a} \times(\vec{b} \times \vec{c})+\vec{b} \times(\vec{c} \times \vec{a})+\vec{c} \times(\vec{a} \times \vec{b})=\overrightarrow{0}\)