268973
A vector parallel to the vector (i j $ + 2$) and having magnitude \(3 \sqrt{5}\) units is
1 \(3 \hat{i}+6 \hat{j}\)
2 \(6 \hat{i}-3 \hat{j}\)
3 \(4 \hat{i}-2 \hat{j}\)
4 \(\hat{i}-2 \hat{j}\)
Explanation:
\(\quad \vec{B}=K \frac{\vec{A}}{|\vec{A}|}\)
VECTORS
268974
If \(\vec{A}=5 \hat{i}-2 \hat{j}+3 \hat{k}\) and \(\vec{B}=2 \hat{i}+\hat{j}+2 \hat{k}\), component of \(\vec{B}\) along \(\vec{A}\) is
1 \(\frac{\sqrt{14}}{38}\)
2 \(\frac{28}{\sqrt{38}}\)
3 \(\frac{\sqrt{28}}{38}\)
4 \(\frac{14}{\sqrt{38}}\)
Explanation:
\(b \cos \theta=\frac{\vec{a} \cdot \vec{b}}{a}\)
VECTORS
268975
If the vectors \(\vec{A}=a \hat{i}+\hat{j}-2 \hat{k}\) and \(\vec{B}=a \hat{i}-a \hat{j}+\hat{k}\) are perpendicular to each other then the positive value of ' \(a\) ' is
268976
When a force \((8 \hat{i}+4 \hat{j})\) newton displaces a particle through \((3 \hat{i}-3 \hat{j})\) metre, the power is \(0.6 \mathrm{~W}\). The time of action of the force is
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VECTORS
268973
A vector parallel to the vector (i j $ + 2$) and having magnitude \(3 \sqrt{5}\) units is
1 \(3 \hat{i}+6 \hat{j}\)
2 \(6 \hat{i}-3 \hat{j}\)
3 \(4 \hat{i}-2 \hat{j}\)
4 \(\hat{i}-2 \hat{j}\)
Explanation:
\(\quad \vec{B}=K \frac{\vec{A}}{|\vec{A}|}\)
VECTORS
268974
If \(\vec{A}=5 \hat{i}-2 \hat{j}+3 \hat{k}\) and \(\vec{B}=2 \hat{i}+\hat{j}+2 \hat{k}\), component of \(\vec{B}\) along \(\vec{A}\) is
1 \(\frac{\sqrt{14}}{38}\)
2 \(\frac{28}{\sqrt{38}}\)
3 \(\frac{\sqrt{28}}{38}\)
4 \(\frac{14}{\sqrt{38}}\)
Explanation:
\(b \cos \theta=\frac{\vec{a} \cdot \vec{b}}{a}\)
VECTORS
268975
If the vectors \(\vec{A}=a \hat{i}+\hat{j}-2 \hat{k}\) and \(\vec{B}=a \hat{i}-a \hat{j}+\hat{k}\) are perpendicular to each other then the positive value of ' \(a\) ' is
268976
When a force \((8 \hat{i}+4 \hat{j})\) newton displaces a particle through \((3 \hat{i}-3 \hat{j})\) metre, the power is \(0.6 \mathrm{~W}\). The time of action of the force is
268973
A vector parallel to the vector (i j $ + 2$) and having magnitude \(3 \sqrt{5}\) units is
1 \(3 \hat{i}+6 \hat{j}\)
2 \(6 \hat{i}-3 \hat{j}\)
3 \(4 \hat{i}-2 \hat{j}\)
4 \(\hat{i}-2 \hat{j}\)
Explanation:
\(\quad \vec{B}=K \frac{\vec{A}}{|\vec{A}|}\)
VECTORS
268974
If \(\vec{A}=5 \hat{i}-2 \hat{j}+3 \hat{k}\) and \(\vec{B}=2 \hat{i}+\hat{j}+2 \hat{k}\), component of \(\vec{B}\) along \(\vec{A}\) is
1 \(\frac{\sqrt{14}}{38}\)
2 \(\frac{28}{\sqrt{38}}\)
3 \(\frac{\sqrt{28}}{38}\)
4 \(\frac{14}{\sqrt{38}}\)
Explanation:
\(b \cos \theta=\frac{\vec{a} \cdot \vec{b}}{a}\)
VECTORS
268975
If the vectors \(\vec{A}=a \hat{i}+\hat{j}-2 \hat{k}\) and \(\vec{B}=a \hat{i}-a \hat{j}+\hat{k}\) are perpendicular to each other then the positive value of ' \(a\) ' is
268976
When a force \((8 \hat{i}+4 \hat{j})\) newton displaces a particle through \((3 \hat{i}-3 \hat{j})\) metre, the power is \(0.6 \mathrm{~W}\). The time of action of the force is
268973
A vector parallel to the vector (i j $ + 2$) and having magnitude \(3 \sqrt{5}\) units is
1 \(3 \hat{i}+6 \hat{j}\)
2 \(6 \hat{i}-3 \hat{j}\)
3 \(4 \hat{i}-2 \hat{j}\)
4 \(\hat{i}-2 \hat{j}\)
Explanation:
\(\quad \vec{B}=K \frac{\vec{A}}{|\vec{A}|}\)
VECTORS
268974
If \(\vec{A}=5 \hat{i}-2 \hat{j}+3 \hat{k}\) and \(\vec{B}=2 \hat{i}+\hat{j}+2 \hat{k}\), component of \(\vec{B}\) along \(\vec{A}\) is
1 \(\frac{\sqrt{14}}{38}\)
2 \(\frac{28}{\sqrt{38}}\)
3 \(\frac{\sqrt{28}}{38}\)
4 \(\frac{14}{\sqrt{38}}\)
Explanation:
\(b \cos \theta=\frac{\vec{a} \cdot \vec{b}}{a}\)
VECTORS
268975
If the vectors \(\vec{A}=a \hat{i}+\hat{j}-2 \hat{k}\) and \(\vec{B}=a \hat{i}-a \hat{j}+\hat{k}\) are perpendicular to each other then the positive value of ' \(a\) ' is
268976
When a force \((8 \hat{i}+4 \hat{j})\) newton displaces a particle through \((3 \hat{i}-3 \hat{j})\) metre, the power is \(0.6 \mathrm{~W}\). The time of action of the force is