ADDITION,SUBTRACTIONAND RESOLUTION OF VECTORS
VECTORS

268999 A particle has a displacement of\(\mathbf{1 2} \mathrm{m}\) towards east then \(5 \mathrm{~m}\) towards north and then \(6 \mathrm{~m}\) vertically upwards the resultant displacement is nearly

1 \(10.04 \mathrm{~m}\)
2 \(12.10 \mathrm{~m}\)
3 \(14.32 \mathrm{~m}\)
4 \(13.06 \mathrm{~m}\)
VECTORS

269001 \(O\) is a point on the ground chosen as origin. \(A\) body first suffers a displacement of \(10 \sqrt{2} \mathbf{m}\) North - East, next \(10 \mathrm{~m}\) north and finally \(10 \sqrt{2}\) North-West. How far it is from the origin ?

1 \(30 \mathrm{~m}\) north
2 \(30 \mathrm{~m}\) south
3 \(30 \mathrm{~m}\) west
4 \(30 \mathrm{~m}\) east
VECTORS

269002 If the two directional cosines of a vectors are\(\frac{1}{\sqrt{2}}\) and \(\frac{1}{\sqrt{3}}\) then the value of third directional cosine is

1 \(\frac{1}{\sqrt{6}}\)
2 \(\frac{1}{\sqrt{5}}\)
3 \(\frac{1}{\sqrt{7}}\)
4 \(\frac{1}{\sqrt{10}}\)
VECTORS

269000 Four co-planar concurrent forces are acting on a body as shown in the figure to keep it in equilibrium. Then the values of\(P\) and \(\theta\) are.

1 \(P=4 N, \theta=0^{0}\)
2 \(\mathrm{P}=2 \mathrm{~N}, \theta=90^{\circ}\)
3 \(\mathrm{P}=2 \mathrm{~N}, \theta=0^{0}\)
4 \(\mathrm{P}=4 \mathrm{~N}, \theta=90^{\circ}\)
VECTORS

268999 A particle has a displacement of\(\mathbf{1 2} \mathrm{m}\) towards east then \(5 \mathrm{~m}\) towards north and then \(6 \mathrm{~m}\) vertically upwards the resultant displacement is nearly

1 \(10.04 \mathrm{~m}\)
2 \(12.10 \mathrm{~m}\)
3 \(14.32 \mathrm{~m}\)
4 \(13.06 \mathrm{~m}\)
VECTORS

269001 \(O\) is a point on the ground chosen as origin. \(A\) body first suffers a displacement of \(10 \sqrt{2} \mathbf{m}\) North - East, next \(10 \mathrm{~m}\) north and finally \(10 \sqrt{2}\) North-West. How far it is from the origin ?

1 \(30 \mathrm{~m}\) north
2 \(30 \mathrm{~m}\) south
3 \(30 \mathrm{~m}\) west
4 \(30 \mathrm{~m}\) east
VECTORS

269002 If the two directional cosines of a vectors are\(\frac{1}{\sqrt{2}}\) and \(\frac{1}{\sqrt{3}}\) then the value of third directional cosine is

1 \(\frac{1}{\sqrt{6}}\)
2 \(\frac{1}{\sqrt{5}}\)
3 \(\frac{1}{\sqrt{7}}\)
4 \(\frac{1}{\sqrt{10}}\)
VECTORS

269000 Four co-planar concurrent forces are acting on a body as shown in the figure to keep it in equilibrium. Then the values of\(P\) and \(\theta\) are.

1 \(P=4 N, \theta=0^{0}\)
2 \(\mathrm{P}=2 \mathrm{~N}, \theta=90^{\circ}\)
3 \(\mathrm{P}=2 \mathrm{~N}, \theta=0^{0}\)
4 \(\mathrm{P}=4 \mathrm{~N}, \theta=90^{\circ}\)
VECTORS

268999 A particle has a displacement of\(\mathbf{1 2} \mathrm{m}\) towards east then \(5 \mathrm{~m}\) towards north and then \(6 \mathrm{~m}\) vertically upwards the resultant displacement is nearly

1 \(10.04 \mathrm{~m}\)
2 \(12.10 \mathrm{~m}\)
3 \(14.32 \mathrm{~m}\)
4 \(13.06 \mathrm{~m}\)
VECTORS

269001 \(O\) is a point on the ground chosen as origin. \(A\) body first suffers a displacement of \(10 \sqrt{2} \mathbf{m}\) North - East, next \(10 \mathrm{~m}\) north and finally \(10 \sqrt{2}\) North-West. How far it is from the origin ?

1 \(30 \mathrm{~m}\) north
2 \(30 \mathrm{~m}\) south
3 \(30 \mathrm{~m}\) west
4 \(30 \mathrm{~m}\) east
VECTORS

269002 If the two directional cosines of a vectors are\(\frac{1}{\sqrt{2}}\) and \(\frac{1}{\sqrt{3}}\) then the value of third directional cosine is

1 \(\frac{1}{\sqrt{6}}\)
2 \(\frac{1}{\sqrt{5}}\)
3 \(\frac{1}{\sqrt{7}}\)
4 \(\frac{1}{\sqrt{10}}\)
VECTORS

269000 Four co-planar concurrent forces are acting on a body as shown in the figure to keep it in equilibrium. Then the values of\(P\) and \(\theta\) are.

1 \(P=4 N, \theta=0^{0}\)
2 \(\mathrm{P}=2 \mathrm{~N}, \theta=90^{\circ}\)
3 \(\mathrm{P}=2 \mathrm{~N}, \theta=0^{0}\)
4 \(\mathrm{P}=4 \mathrm{~N}, \theta=90^{\circ}\)
NEET Test Series from KOTA - 10 Papers In MS WORD WhatsApp Here
VECTORS

268999 A particle has a displacement of\(\mathbf{1 2} \mathrm{m}\) towards east then \(5 \mathrm{~m}\) towards north and then \(6 \mathrm{~m}\) vertically upwards the resultant displacement is nearly

1 \(10.04 \mathrm{~m}\)
2 \(12.10 \mathrm{~m}\)
3 \(14.32 \mathrm{~m}\)
4 \(13.06 \mathrm{~m}\)
VECTORS

269001 \(O\) is a point on the ground chosen as origin. \(A\) body first suffers a displacement of \(10 \sqrt{2} \mathbf{m}\) North - East, next \(10 \mathrm{~m}\) north and finally \(10 \sqrt{2}\) North-West. How far it is from the origin ?

1 \(30 \mathrm{~m}\) north
2 \(30 \mathrm{~m}\) south
3 \(30 \mathrm{~m}\) west
4 \(30 \mathrm{~m}\) east
VECTORS

269002 If the two directional cosines of a vectors are\(\frac{1}{\sqrt{2}}\) and \(\frac{1}{\sqrt{3}}\) then the value of third directional cosine is

1 \(\frac{1}{\sqrt{6}}\)
2 \(\frac{1}{\sqrt{5}}\)
3 \(\frac{1}{\sqrt{7}}\)
4 \(\frac{1}{\sqrt{10}}\)
VECTORS

269000 Four co-planar concurrent forces are acting on a body as shown in the figure to keep it in equilibrium. Then the values of\(P\) and \(\theta\) are.

1 \(P=4 N, \theta=0^{0}\)
2 \(\mathrm{P}=2 \mathrm{~N}, \theta=90^{\circ}\)
3 \(\mathrm{P}=2 \mathrm{~N}, \theta=0^{0}\)
4 \(\mathrm{P}=4 \mathrm{~N}, \theta=90^{\circ}\)