355534
If \(|\vec{A}-\vec{B}|=|\vec{A}|-|\vec{B}|\), the angle between \(\vec{A}\) and \(\vec{B}\) is
1 \(60^{\circ}\)
2 \(0^{\circ}\)
3 \(120^{\circ}\)
4 \(90^{\circ}\)
Explanation:
\(|\vec A - \vec B| = \sqrt {{A^2} + {B^2} - 2AB\cos \theta } \) Given that \(\sqrt{A^{2}+B^{2}-2 A B \cos \theta}=A-B\) Squaring on both sides \(\begin{aligned}& A^{2}+B^{2}-2 A B \cos \theta=A^{2}+B^{2}-2 A B \\& \cos \theta=1 \Rightarrow \theta=0^{\circ}\end{aligned}\)
PHXI06:WORK ENERGY AND POWER
355535
If for two vectors \(\vec{A}\) and \(\vec{B}\), sum \((\vec{A}+\vec{B})\) is perpendicular to the difference \((\vec{A}-\vec{B})\). The ratio of their magnitudes is
1 1
2 2
3 3
4 None of these
Explanation:
\((\vec{A}+\vec{B})\) is perpendicular to \((\vec{A}-\vec{B})\). Thus \((\vec{A}+\vec{B}) \cdot(\vec{A}-\vec{B})=0\) or \(A^{2}+\vec{B} \cdot \vec{A}-\vec{A} \cdot \vec{B}-B^{2}=0\). Because of commutative of dot product \(\begin{aligned}& \vec{A} \cdot \vec{B}=\vec{B} \cdot \vec{A} \\& \therefore A^{2}-B^{2}=0 \text { or } A=B\end{aligned}\) Thus the ratio of magnitudes \(A / B=1\)
PHXI06:WORK ENERGY AND POWER
355536
If \({\vec{a}, \vec{b}}\) and \({\vec{c}}\) are non-zero coplanar vectors, such that \({\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{c}=0}\). If \({|\vec{a}|=1}\) unit and \({|\vec{c}|=6}\) units. The value of \({|\vec{a} \cdot \vec{c}|}\). Is
1 3
2 9
3 6
4 5
Explanation:
\({\because \vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{c}=0}\) It means \({\vec{a}}\) and \({\vec{b}}\) are perpendicular to each other. Similarly, \({\ddot{b}}\) and \({\vec{c}}\) are perpendicular to each other. So, we can say that \({\vec{a}}\) and \({\vec{c}}\) are parallel to each other \(\therefore \vec{a} \cdot \vec{c}=|a||c| \cos 0^{\circ}=1 \times 6=6 \text { units }\)
PHXI06:WORK ENERGY AND POWER
355537
The angle between the two vectors \(\vec{A}=5 \hat{i}+5 \hat{j}\) and \(\vec{B}=5 \hat{i}-5 \hat{j}\) will be
355534
If \(|\vec{A}-\vec{B}|=|\vec{A}|-|\vec{B}|\), the angle between \(\vec{A}\) and \(\vec{B}\) is
1 \(60^{\circ}\)
2 \(0^{\circ}\)
3 \(120^{\circ}\)
4 \(90^{\circ}\)
Explanation:
\(|\vec A - \vec B| = \sqrt {{A^2} + {B^2} - 2AB\cos \theta } \) Given that \(\sqrt{A^{2}+B^{2}-2 A B \cos \theta}=A-B\) Squaring on both sides \(\begin{aligned}& A^{2}+B^{2}-2 A B \cos \theta=A^{2}+B^{2}-2 A B \\& \cos \theta=1 \Rightarrow \theta=0^{\circ}\end{aligned}\)
PHXI06:WORK ENERGY AND POWER
355535
If for two vectors \(\vec{A}\) and \(\vec{B}\), sum \((\vec{A}+\vec{B})\) is perpendicular to the difference \((\vec{A}-\vec{B})\). The ratio of their magnitudes is
1 1
2 2
3 3
4 None of these
Explanation:
\((\vec{A}+\vec{B})\) is perpendicular to \((\vec{A}-\vec{B})\). Thus \((\vec{A}+\vec{B}) \cdot(\vec{A}-\vec{B})=0\) or \(A^{2}+\vec{B} \cdot \vec{A}-\vec{A} \cdot \vec{B}-B^{2}=0\). Because of commutative of dot product \(\begin{aligned}& \vec{A} \cdot \vec{B}=\vec{B} \cdot \vec{A} \\& \therefore A^{2}-B^{2}=0 \text { or } A=B\end{aligned}\) Thus the ratio of magnitudes \(A / B=1\)
PHXI06:WORK ENERGY AND POWER
355536
If \({\vec{a}, \vec{b}}\) and \({\vec{c}}\) are non-zero coplanar vectors, such that \({\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{c}=0}\). If \({|\vec{a}|=1}\) unit and \({|\vec{c}|=6}\) units. The value of \({|\vec{a} \cdot \vec{c}|}\). Is
1 3
2 9
3 6
4 5
Explanation:
\({\because \vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{c}=0}\) It means \({\vec{a}}\) and \({\vec{b}}\) are perpendicular to each other. Similarly, \({\ddot{b}}\) and \({\vec{c}}\) are perpendicular to each other. So, we can say that \({\vec{a}}\) and \({\vec{c}}\) are parallel to each other \(\therefore \vec{a} \cdot \vec{c}=|a||c| \cos 0^{\circ}=1 \times 6=6 \text { units }\)
PHXI06:WORK ENERGY AND POWER
355537
The angle between the two vectors \(\vec{A}=5 \hat{i}+5 \hat{j}\) and \(\vec{B}=5 \hat{i}-5 \hat{j}\) will be
355534
If \(|\vec{A}-\vec{B}|=|\vec{A}|-|\vec{B}|\), the angle between \(\vec{A}\) and \(\vec{B}\) is
1 \(60^{\circ}\)
2 \(0^{\circ}\)
3 \(120^{\circ}\)
4 \(90^{\circ}\)
Explanation:
\(|\vec A - \vec B| = \sqrt {{A^2} + {B^2} - 2AB\cos \theta } \) Given that \(\sqrt{A^{2}+B^{2}-2 A B \cos \theta}=A-B\) Squaring on both sides \(\begin{aligned}& A^{2}+B^{2}-2 A B \cos \theta=A^{2}+B^{2}-2 A B \\& \cos \theta=1 \Rightarrow \theta=0^{\circ}\end{aligned}\)
PHXI06:WORK ENERGY AND POWER
355535
If for two vectors \(\vec{A}\) and \(\vec{B}\), sum \((\vec{A}+\vec{B})\) is perpendicular to the difference \((\vec{A}-\vec{B})\). The ratio of their magnitudes is
1 1
2 2
3 3
4 None of these
Explanation:
\((\vec{A}+\vec{B})\) is perpendicular to \((\vec{A}-\vec{B})\). Thus \((\vec{A}+\vec{B}) \cdot(\vec{A}-\vec{B})=0\) or \(A^{2}+\vec{B} \cdot \vec{A}-\vec{A} \cdot \vec{B}-B^{2}=0\). Because of commutative of dot product \(\begin{aligned}& \vec{A} \cdot \vec{B}=\vec{B} \cdot \vec{A} \\& \therefore A^{2}-B^{2}=0 \text { or } A=B\end{aligned}\) Thus the ratio of magnitudes \(A / B=1\)
PHXI06:WORK ENERGY AND POWER
355536
If \({\vec{a}, \vec{b}}\) and \({\vec{c}}\) are non-zero coplanar vectors, such that \({\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{c}=0}\). If \({|\vec{a}|=1}\) unit and \({|\vec{c}|=6}\) units. The value of \({|\vec{a} \cdot \vec{c}|}\). Is
1 3
2 9
3 6
4 5
Explanation:
\({\because \vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{c}=0}\) It means \({\vec{a}}\) and \({\vec{b}}\) are perpendicular to each other. Similarly, \({\ddot{b}}\) and \({\vec{c}}\) are perpendicular to each other. So, we can say that \({\vec{a}}\) and \({\vec{c}}\) are parallel to each other \(\therefore \vec{a} \cdot \vec{c}=|a||c| \cos 0^{\circ}=1 \times 6=6 \text { units }\)
PHXI06:WORK ENERGY AND POWER
355537
The angle between the two vectors \(\vec{A}=5 \hat{i}+5 \hat{j}\) and \(\vec{B}=5 \hat{i}-5 \hat{j}\) will be
355534
If \(|\vec{A}-\vec{B}|=|\vec{A}|-|\vec{B}|\), the angle between \(\vec{A}\) and \(\vec{B}\) is
1 \(60^{\circ}\)
2 \(0^{\circ}\)
3 \(120^{\circ}\)
4 \(90^{\circ}\)
Explanation:
\(|\vec A - \vec B| = \sqrt {{A^2} + {B^2} - 2AB\cos \theta } \) Given that \(\sqrt{A^{2}+B^{2}-2 A B \cos \theta}=A-B\) Squaring on both sides \(\begin{aligned}& A^{2}+B^{2}-2 A B \cos \theta=A^{2}+B^{2}-2 A B \\& \cos \theta=1 \Rightarrow \theta=0^{\circ}\end{aligned}\)
PHXI06:WORK ENERGY AND POWER
355535
If for two vectors \(\vec{A}\) and \(\vec{B}\), sum \((\vec{A}+\vec{B})\) is perpendicular to the difference \((\vec{A}-\vec{B})\). The ratio of their magnitudes is
1 1
2 2
3 3
4 None of these
Explanation:
\((\vec{A}+\vec{B})\) is perpendicular to \((\vec{A}-\vec{B})\). Thus \((\vec{A}+\vec{B}) \cdot(\vec{A}-\vec{B})=0\) or \(A^{2}+\vec{B} \cdot \vec{A}-\vec{A} \cdot \vec{B}-B^{2}=0\). Because of commutative of dot product \(\begin{aligned}& \vec{A} \cdot \vec{B}=\vec{B} \cdot \vec{A} \\& \therefore A^{2}-B^{2}=0 \text { or } A=B\end{aligned}\) Thus the ratio of magnitudes \(A / B=1\)
PHXI06:WORK ENERGY AND POWER
355536
If \({\vec{a}, \vec{b}}\) and \({\vec{c}}\) are non-zero coplanar vectors, such that \({\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{c}=0}\). If \({|\vec{a}|=1}\) unit and \({|\vec{c}|=6}\) units. The value of \({|\vec{a} \cdot \vec{c}|}\). Is
1 3
2 9
3 6
4 5
Explanation:
\({\because \vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{c}=0}\) It means \({\vec{a}}\) and \({\vec{b}}\) are perpendicular to each other. Similarly, \({\ddot{b}}\) and \({\vec{c}}\) are perpendicular to each other. So, we can say that \({\vec{a}}\) and \({\vec{c}}\) are parallel to each other \(\therefore \vec{a} \cdot \vec{c}=|a||c| \cos 0^{\circ}=1 \times 6=6 \text { units }\)
PHXI06:WORK ENERGY AND POWER
355537
The angle between the two vectors \(\vec{A}=5 \hat{i}+5 \hat{j}\) and \(\vec{B}=5 \hat{i}-5 \hat{j}\) will be
