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Question

Establish the following vector inequalities geometrically or otherwise : (a) |a+b| < |a| + |b| (b) |a+b| > ||a| −|b|| (c) |a−b| < |a| + |b| (d) |a−b| > ||a| − |b|| When does the equality sign above apply?

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Solution

a)

The figure below shoes the parallelogram, whose adjacent sides are two vectors, say a and b, the diagonal of the parallelogram represents the vector a+b.



From parallelogram OMNP,

OM=PN=| a | OP=MN=| b | ON=| a+b |

From properties of triangle, the length of each side of triangle is less than the sum of other two sides of the triangle.

In ΔOMN,

ON<( OM+ON )

Substitute the values in the above expression.

| a+b |<| a |+| b |(I)

If a and b acts along the straight line in same direction, then

| a+b |=| a |+| b |(II)

From equation (I) and equation (II),

| a+b || a |+| b |

Thus, it is proved that | a+b || a |+| b |.

b)

The figure below shoes the parallelogram, whose adjacent sides are two vectors, say a and b, the diagonal of the parallelogram represents the vector a+b.



From parallelogram OMNP,

OM=PN=| a | OP=MN=| b | ON=| a+b |

From properties of triangle, the length of each side of triangle is less than the sum of other two sides of the triangle.

In ΔOMN,

ON+OM>MN | ON |>| MNOM |

Substitute the values in the above expression.

| | a+b | |>| | a || b | | | a+b |>| | a || b | | (III)

If a and b acts along the straight line in same direction, then

| a+b |=| | a || b | |(IV)

From equation (III) and equation (IV),

| a+b || | a || b | |

Thus, it is proved that | a+b || | a || b | |.

c)

The figure below shoes the parallelogram, whose adjacent sides are two vectors, say a and b, the diagonal of the parallelogram represents the vector ab.



From parallelogram OPSR,

OR=PS=| b | OP=| a | OS=| ab |

From properties of triangle, the length of each side of triangle is less than the sum of other two sides of the triangle.

In ΔOPS,

OS<OP+PS

Substitute the values in the above expression.

| ab |<| a |+| b |(V)

If a and b acts along the straight line in same direction, then

| a+b |=| | a || b | |(VI)

From equation (V) and equation (VI),

| ab || a |+| b |

Thus, it is proved that | ab || a |+| b |.

d)

The figure below shoes the parallelogram, whose adjacent sides are two vectors, say a and b, the diagonal of the parallelogram represents the vector ab.



From parallelogram OPSR,

OR=PS=| b | OP=| a | OS=| ab |

From properties of triangle, the length of each side of triangle is less than the sum of other two sides of the triangle.

In ΔOPS,

OS+PS>OP OS>OPPS | OS |>| OPPS |

Substitute the values in the above expression.

| | ab | |>| | a || b | | | ab |>| | a || b | | (VII)

If a and b acts along the straight line in same direction, then

| ab |=| | a || b | |(VIII)

From equation (VII) and equation (VIII),

| ab || | a || b | |

Thus, it is proved that | ab || | a || b | |.


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