
For any vector $a$ , which is the value of ${\left| {a \times i} \right|^2} + {\left| {a \times j} \right|^2} + {\left| {a \times k} \right|^2}$ ?
A. ${\left| a \right|^2}$
B. $2{\left| a \right|^2}$
C. $3{\left| a \right|^2}$
D. $4{\left| a \right|^2}$
Answer
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Hint: To find the solution to a given problem, the properties of cross-product (vector-product) between two vectors must be known. In this problem, we will first assume any random vector $a$, apply the properties of a cross-product and then compute the modulus of the resulting vector to find the correct solution to the given problem.
Formula used:
The cross product of unit vectors used in this problem are: -
$\widehat i \times \widehat i = \widehat j \times \widehat j = \widehat k \times \widehat k = 0$ and $\widehat i \times \widehat j = - \widehat j \times \widehat i = \widehat k$ , $\widehat j \times \widehat k = - \widehat k \times \widehat j = \widehat i$ & $\widehat k \times \widehat i = - \widehat i \times \widehat k = \widehat j$and the modulus of any vector $a$ is given as:
$\left| {\overrightarrow a } \right| = \left( {\sqrt {{x^2} + {y^2} + {z^2}} } \right)$
Complete step by step solution:
Let us first consider the value of $\overrightarrow a = x\widehat i + y\widehat j + z\widehat k$.
Let, ${\left| {a \times i} \right|^2} + {\left| {a \times j} \right|^2} + {\left| {a \times k} \right|^2}$ $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\,(1)$
Now, let us find out the values of the given expressions one by one in the problem: -
For cross-product with Vector $\widehat i$ ,
$a \times \widehat i = \left( {(x\widehat i + y\widehat j + z\widehat k) \times \widehat i} \right) = \left( {x(\widehat i \times \widehat i) + y(\widehat j \times \widehat i) + z(\widehat k \times \widehat i)} \right) = \left( {0 + y( - \widehat k) + z(\widehat j)} \right) = \left( { - y\widehat k + z\widehat j} \right)$
By taking the modulus and then squaring it, we will get: -
${\left| {a \times \widehat i} \right|^2} = {\left( {\sqrt {{{( - y)}^2} + {z^2}} } \right)^2} = {y^2} + {z^2}$
For cross-product with Vector $\widehat j$ ,
$a \times \widehat j = \left( {(x\widehat i + y\widehat j + z\widehat k) \times \widehat j} \right) = \left( {x(\widehat i \times \widehat j) + y(\widehat j \times \widehat j) + z(\widehat k \times \widehat j)} \right) = \left( {x(\widehat k) + 0 + z( - \widehat i)} \right) = \left( {x\widehat k - z\widehat i} \right)$By taking the modulus and then squaring it, we will get: -
${\left| {a \times \widehat j} \right|^2} = {\left( {\sqrt {{x^2} + {{( - z)}^2}} } \right)^2} = {x^2} + {z^2}$
For cross-product with Vector $\widehat k$ ,
By taking the modulus and then squaring it, we will get: -
$a \times \widehat k = \left( {(x\widehat i + y\widehat j + z\widehat k) \times \widehat k} \right) = \left( {x(\widehat i \times \widehat k) + y(\widehat j \times \widehat k) + z(\widehat k \times \widehat k)} \right) = \left( {x( - \widehat j) + y(\widehat i) + 0} \right) = \left( { - x\widehat j + y\widehat i} \right)$ -
${\left| {a \times \widehat k} \right|^2} = {\left( {\sqrt {{{( - x)}^2} + {y^2}} } \right)^2} = {x^2} + {y^2}$
Now substitute the values in eq. $\left( 1 \right)$ , we get
${\left| {a \times i} \right|^2} + {\left| {a \times j} \right|^2} + {\left| {a \times k} \right|^2} = ({y^2} + {z^2}) + ({x^2} + {z^2}) + ({x^2} + {y^2}) = 2({x^2} + {y^2} + {z^2})$ $\,\,...\,(2)$
Let us find the modulus of the vector $a$and square it, we get
${\left| {\overrightarrow a } \right|^2} = {\left( {\sqrt {{x^2} + {y^2} + {z^2}} } \right)^2} = {x^2} + {y^2} + {z^2}$
Now, on substituting this value in equation $\left( 2 \right)$ , we get
${\left| {a \times i} \right|^2} + {\left| {a \times j} \right|^2} + {\left| {a \times k} \right|^2} = 2({x^2} + {y^2} + {z^2}) = 2{\left| {\overrightarrow a } \right|^2}$
Thus, the final value of the expression given is $2{\left| {\overrightarrow a } \right|^2}$.
Hence, the correct option is: (B) $2{\left| a \right|^2}$
Note: Since the problem is based on Vector Algebra, it is essential to analyze the given conditions carefully on the basis of which the procedure of solving the problem is identified. Also, we can use the natural formula of cross-product by using the determinant method here. Calculations must be performed very carefully and all the signs of the vector must be used in the solution in a proper manner.
Formula used:
The cross product of unit vectors used in this problem are: -
$\widehat i \times \widehat i = \widehat j \times \widehat j = \widehat k \times \widehat k = 0$ and $\widehat i \times \widehat j = - \widehat j \times \widehat i = \widehat k$ , $\widehat j \times \widehat k = - \widehat k \times \widehat j = \widehat i$ & $\widehat k \times \widehat i = - \widehat i \times \widehat k = \widehat j$and the modulus of any vector $a$ is given as:
$\left| {\overrightarrow a } \right| = \left( {\sqrt {{x^2} + {y^2} + {z^2}} } \right)$
Complete step by step solution:
Let us first consider the value of $\overrightarrow a = x\widehat i + y\widehat j + z\widehat k$.
Let, ${\left| {a \times i} \right|^2} + {\left| {a \times j} \right|^2} + {\left| {a \times k} \right|^2}$ $\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\,(1)$
Now, let us find out the values of the given expressions one by one in the problem: -
For cross-product with Vector $\widehat i$ ,
$a \times \widehat i = \left( {(x\widehat i + y\widehat j + z\widehat k) \times \widehat i} \right) = \left( {x(\widehat i \times \widehat i) + y(\widehat j \times \widehat i) + z(\widehat k \times \widehat i)} \right) = \left( {0 + y( - \widehat k) + z(\widehat j)} \right) = \left( { - y\widehat k + z\widehat j} \right)$
By taking the modulus and then squaring it, we will get: -
${\left| {a \times \widehat i} \right|^2} = {\left( {\sqrt {{{( - y)}^2} + {z^2}} } \right)^2} = {y^2} + {z^2}$
For cross-product with Vector $\widehat j$ ,
$a \times \widehat j = \left( {(x\widehat i + y\widehat j + z\widehat k) \times \widehat j} \right) = \left( {x(\widehat i \times \widehat j) + y(\widehat j \times \widehat j) + z(\widehat k \times \widehat j)} \right) = \left( {x(\widehat k) + 0 + z( - \widehat i)} \right) = \left( {x\widehat k - z\widehat i} \right)$By taking the modulus and then squaring it, we will get: -
${\left| {a \times \widehat j} \right|^2} = {\left( {\sqrt {{x^2} + {{( - z)}^2}} } \right)^2} = {x^2} + {z^2}$
For cross-product with Vector $\widehat k$ ,
By taking the modulus and then squaring it, we will get: -
$a \times \widehat k = \left( {(x\widehat i + y\widehat j + z\widehat k) \times \widehat k} \right) = \left( {x(\widehat i \times \widehat k) + y(\widehat j \times \widehat k) + z(\widehat k \times \widehat k)} \right) = \left( {x( - \widehat j) + y(\widehat i) + 0} \right) = \left( { - x\widehat j + y\widehat i} \right)$ -
${\left| {a \times \widehat k} \right|^2} = {\left( {\sqrt {{{( - x)}^2} + {y^2}} } \right)^2} = {x^2} + {y^2}$
Now substitute the values in eq. $\left( 1 \right)$ , we get
${\left| {a \times i} \right|^2} + {\left| {a \times j} \right|^2} + {\left| {a \times k} \right|^2} = ({y^2} + {z^2}) + ({x^2} + {z^2}) + ({x^2} + {y^2}) = 2({x^2} + {y^2} + {z^2})$ $\,\,...\,(2)$
Let us find the modulus of the vector $a$and square it, we get
${\left| {\overrightarrow a } \right|^2} = {\left( {\sqrt {{x^2} + {y^2} + {z^2}} } \right)^2} = {x^2} + {y^2} + {z^2}$
Now, on substituting this value in equation $\left( 2 \right)$ , we get
${\left| {a \times i} \right|^2} + {\left| {a \times j} \right|^2} + {\left| {a \times k} \right|^2} = 2({x^2} + {y^2} + {z^2}) = 2{\left| {\overrightarrow a } \right|^2}$
Thus, the final value of the expression given is $2{\left| {\overrightarrow a } \right|^2}$.
Hence, the correct option is: (B) $2{\left| a \right|^2}$
Note: Since the problem is based on Vector Algebra, it is essential to analyze the given conditions carefully on the basis of which the procedure of solving the problem is identified. Also, we can use the natural formula of cross-product by using the determinant method here. Calculations must be performed very carefully and all the signs of the vector must be used in the solution in a proper manner.
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