
One side of the parallelogram has length 3 and another side has length 4. Let a and b denote the lengths of the diagonals of the parallelogram. Which of the following quantities can be determined from the given information?
${\text{(i) }}a + b$
${\text{(ii) }}{a^2} + {b^2}{\text{ }}$
${\text{(iii) }}{a^3} + {b^3}$
A) Only (i)
B) Only (ii)
C) Only (iii)
D) Only (i) and (ii)
Answer
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Hint: Two vectors $\vec a$ and $\vec b$represented by the two adjacent sides of a parallelogram in magnitude and direction, then their sum $\vec a + \vec b$ is represented in magnitude and direction by the diagonal of the parallelogram through their common path. This is known as the parallelogram law of vector addition.
The dot product or scalar product of two vectors $\vec a{\text{ and }}\vec b$ is the multiplication of the magnitude of two vectors and cos of the angle between them.
$\vec a \cdot \vec b = \left| {\vec a} \right|\left| {\vec b} \right|\cos \theta $,
$\vec a \cdot \vec a = {\left| {\vec a} \right|^2}$ [$\because \cos 0^\circ = 1$] …………………………...…… (1)
Complete step by step answer:
In parallelogram ABCD, AB || CD, AD || BC …… (from figure 2)
and vectors of parallelogram.
$\overrightarrow {AB} = \overrightarrow {DC} ,\overrightarrow {AD} = \overrightarrow {BC} $ …… (from figure 3)
$\overrightarrow {{\text{AC}}} {\text{ }} = {{ \vec a}},\overrightarrow {{\text{ BD}}} {\text{ }} = {{ \vec b}}$, ( $\because $ given) …… (2)
$\Rightarrow \overrightarrow {{\text{AB}}} {\text{ }} + {\text{ }}\overrightarrow {{\text{AD}}} {\text{ }} = {\text{ }}\overrightarrow {{\text{AC}}} $ ( $\because $ parallelogram law of vector addition)
Taking dot product on both sides
${\text{(}}\overrightarrow {{\text{AB}}} {\text{ + }}\overrightarrow {{\text{AD}}} {\text{)}} \cdot {\text{(}}\overrightarrow {{\text{AB}}} {\text{ + }}\overrightarrow {{\text{AD}}} {\text{) = }}\overrightarrow {{\text{AC}}} \cdot \overrightarrow {{\text{AC}}} $
${\left| {\overrightarrow {AC} } \right|^2} = {\left| {\overrightarrow {AB} } \right|^2} + {\left| {\overrightarrow {AD} } \right|^2} + 2\left( {\overrightarrow {AB} .\overrightarrow {AD} } \right)$ (from (1)) …… (3)
$\Rightarrow \overrightarrow {{\text{AB}}} {\text{ }} - {\text{ }}\overrightarrow {{\text{AD}}} {\text{ = }}\overrightarrow {{\text{BD}}} $ ( $\because $ parallelogram law of vector addition)
Taking dot product on both sides
$\Rightarrow {\text{(}}\overrightarrow {{\text{AB}}} - \overrightarrow {{\text{AD}}} {\text{)}} \cdot {\text{(}}\overrightarrow {{\text{AB}}} - \overrightarrow {{\text{AD}}} {\text{) = }}\overrightarrow {{\text{BD}}} \cdot \overrightarrow {{\text{BD}}} $
$\Rightarrow {\left| {\overrightarrow {BD} } \right|^2} = {\left| {\overrightarrow {AB} } \right|^2} + {\left| {\overrightarrow {AD} } \right|^2} - 2\left( {\overrightarrow {AB} .\overrightarrow {AD} } \right)$ …… (4)
Adding equations (2) and (3)
$\Rightarrow {\left| {\overrightarrow {AC} } \right|^2} + {\left| {\overrightarrow {BD} } \right|^2} = 2\left( {{{\left| {\overrightarrow {AB} } \right|}^2} + {{\left| {\overrightarrow {AD} } \right|}^2}} \right)$ …… (5)
Magnitudes: $\left| {\overrightarrow {AC} } \right| = a;{\text{ }}\left| {\overrightarrow {BD} } \right| = b;{\text{ }}\left| {\overrightarrow {AB} } \right| = 3;{\text{ }}\left| {\overrightarrow {AD} } \right| = 4;$ ( $\because $ given)
$\mathop {\text{a}}\nolimits^{\text{2}} {\text{ + }}\mathop {{\text{ b}}}\nolimits^{\text{2}} {\text{ = 2(}}\mathop {\text{3}}\nolimits^{\text{2}} {\text{ + }}\mathop {\text{4}}\nolimits^{\text{2}} {\text{)}}$
$ = 2(9 + 16)$
$ = 2(25)$
$ = 50$
$\therefore$ Option (B): only (ii) is correct. One can only find the value of $\mathop {\text{a}}\nolimits^{\text{2}} {\text{ + }}\mathop {{\text{b }}}\nolimits^{\text{2}} {\text{ = 50}}$.
Note:
One can remember the result derived in the above solution: The sum of squares of length of diagonals of a parallelogram is equal to the sum of squares of length of all the sides of the parallelogram. The result can help solve such problems.
The dot product or scalar product of two vectors $\vec a{\text{ and }}\vec b$ is the multiplication of the magnitude of two vectors and cos of the angle between them.
$\vec a \cdot \vec b = \left| {\vec a} \right|\left| {\vec b} \right|\cos \theta $,
$\vec a \cdot \vec a = {\left| {\vec a} \right|^2}$ [$\because \cos 0^\circ = 1$] …………………………...…… (1)
Complete step by step answer:
In parallelogram ABCD, AB || CD, AD || BC …… (from figure 2)
and vectors of parallelogram.
$\overrightarrow {AB} = \overrightarrow {DC} ,\overrightarrow {AD} = \overrightarrow {BC} $ …… (from figure 3)
$\overrightarrow {{\text{AC}}} {\text{ }} = {{ \vec a}},\overrightarrow {{\text{ BD}}} {\text{ }} = {{ \vec b}}$, ( $\because $ given) …… (2)
$\Rightarrow \overrightarrow {{\text{AB}}} {\text{ }} + {\text{ }}\overrightarrow {{\text{AD}}} {\text{ }} = {\text{ }}\overrightarrow {{\text{AC}}} $ ( $\because $ parallelogram law of vector addition)
Taking dot product on both sides
${\text{(}}\overrightarrow {{\text{AB}}} {\text{ + }}\overrightarrow {{\text{AD}}} {\text{)}} \cdot {\text{(}}\overrightarrow {{\text{AB}}} {\text{ + }}\overrightarrow {{\text{AD}}} {\text{) = }}\overrightarrow {{\text{AC}}} \cdot \overrightarrow {{\text{AC}}} $
${\left| {\overrightarrow {AC} } \right|^2} = {\left| {\overrightarrow {AB} } \right|^2} + {\left| {\overrightarrow {AD} } \right|^2} + 2\left( {\overrightarrow {AB} .\overrightarrow {AD} } \right)$ (from (1)) …… (3)
$\Rightarrow \overrightarrow {{\text{AB}}} {\text{ }} - {\text{ }}\overrightarrow {{\text{AD}}} {\text{ = }}\overrightarrow {{\text{BD}}} $ ( $\because $ parallelogram law of vector addition)
Taking dot product on both sides
$\Rightarrow {\text{(}}\overrightarrow {{\text{AB}}} - \overrightarrow {{\text{AD}}} {\text{)}} \cdot {\text{(}}\overrightarrow {{\text{AB}}} - \overrightarrow {{\text{AD}}} {\text{) = }}\overrightarrow {{\text{BD}}} \cdot \overrightarrow {{\text{BD}}} $
$\Rightarrow {\left| {\overrightarrow {BD} } \right|^2} = {\left| {\overrightarrow {AB} } \right|^2} + {\left| {\overrightarrow {AD} } \right|^2} - 2\left( {\overrightarrow {AB} .\overrightarrow {AD} } \right)$ …… (4)
Adding equations (2) and (3)
$\Rightarrow {\left| {\overrightarrow {AC} } \right|^2} + {\left| {\overrightarrow {BD} } \right|^2} = 2\left( {{{\left| {\overrightarrow {AB} } \right|}^2} + {{\left| {\overrightarrow {AD} } \right|}^2}} \right)$ …… (5)
Magnitudes: $\left| {\overrightarrow {AC} } \right| = a;{\text{ }}\left| {\overrightarrow {BD} } \right| = b;{\text{ }}\left| {\overrightarrow {AB} } \right| = 3;{\text{ }}\left| {\overrightarrow {AD} } \right| = 4;$ ( $\because $ given)
$\mathop {\text{a}}\nolimits^{\text{2}} {\text{ + }}\mathop {{\text{ b}}}\nolimits^{\text{2}} {\text{ = 2(}}\mathop {\text{3}}\nolimits^{\text{2}} {\text{ + }}\mathop {\text{4}}\nolimits^{\text{2}} {\text{)}}$
$ = 2(9 + 16)$
$ = 2(25)$
$ = 50$
$\therefore$ Option (B): only (ii) is correct. One can only find the value of $\mathop {\text{a}}\nolimits^{\text{2}} {\text{ + }}\mathop {{\text{b }}}\nolimits^{\text{2}} {\text{ = 50}}$.
Note:
One can remember the result derived in the above solution: The sum of squares of length of diagonals of a parallelogram is equal to the sum of squares of length of all the sides of the parallelogram. The result can help solve such problems.
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