Question

A vector of length $\sqrt 7$ which is perpendicular to $2\overrightarrow j - \overrightarrow k$ and $- \overrightarrow i + 2\overrightarrow j - 3\overrightarrow k$ makes obtuse angle with $y$-axis is
A) $\dfrac{1}{{\sqrt 5 }}\left( {4\overrightarrow i - \overrightarrow j + \sqrt {18k} } \right)$
B) $\dfrac{1}{{\sqrt 3 }}\left( {4\overrightarrow i - \overrightarrow j - 2\overrightarrow k } \right)$
C) $\dfrac{1}{{\sqrt 3 }}\left( { - 4\overrightarrow i + \overrightarrow j + 2\overrightarrow k } \right)$
D) $\dfrac{1}{{\sqrt 3 }}\left( { - 4\overrightarrow i - \overrightarrow j + 2\overrightarrow k } \right)$

Answer 1

In this problem, using the given data we can find the y-axis. Using vector multiplication and using the formula for find unit vectors we can solve the given problem. The length of the vector is the square root of the sum of the squares of the horizontal and vertical components. The vector product of vectors is perpendicular to them.
The formula used in the problem.
A unit vector
$= \dfrac{{\overrightarrow A \times \overrightarrow B }}{{\left| {\sqrt {{a^2} + {b^2} + {c^2}} } \right|}}$

Complete answer:
In this problem, we are going to solve the problem using given data.
In this problem, two vector equations are given,
The vector equations are,
$\overrightarrow A = 2\overrightarrow j - \overrightarrow k$
$\overrightarrow B = - \overrightarrow i + 2\overrightarrow j - 3\overrightarrow k$
A vector length is the square root of seven which is perpendicular to the given vector equations.
The vector multiplication $\overrightarrow A \times \overrightarrow B = \left( {2\overrightarrow j - \overrightarrow k } \right) \times \left( { - \overrightarrow i + 2\overrightarrow j - 3\overrightarrow k } \right)$
We know that $i \times j = k$ and $j \times i = - k$
$j \times k = i$ and $k \times j = - i$
$k \times i = j$ and $i \times k = - j$
Also, $i \times i = j \times j = k \times k = 0$
Now, we shall do vector multiplication for $\overrightarrow A \times \overrightarrow B = \left( {2\overrightarrow j - \overrightarrow k } \right) \times \left( { - \overrightarrow i + 2\overrightarrow j - 3\overrightarrow k } \right)$
$\overrightarrow A \times \overrightarrow B = \left( {2 \times - 1} \right)\left( {j \times i} \right) + \left( {2 \times 2} \right)\left( {j \times j} \right) + \left( {2 \times - 3} \right)\left( {j \times k} \right) + \left( { - 1 \times - 1} \right)\left( {k \times i} \right) + \left( { - 1 \times 2} \right)\left( {k \times j} \right) + \left( { - 1 \times - 3} \right)\left( {k \times k} \right)$ $\overrightarrow { \Rightarrow A} \times \overrightarrow B = - 2\left( { - k} \right) + 4\left( 0 \right) - 6i + j - 2\left( { - i} \right) + 3\left( 0 \right)$
$\Rightarrow \overrightarrow A \times \overrightarrow B = 2k - 6i + j + 2i$
Now simplify this we get,
$\overrightarrow A \times \overrightarrow B = \left( { - 4\overrightarrow i + \overrightarrow j + 2\overrightarrow k } \right)$
Now we are going to find the unit vector.
Already we know the formula for finding the unit vector.
We can write this as mathematically, $= \dfrac{{\overrightarrow A \times \overrightarrow B }}{{\left| {\sqrt {{a^2} + {b^2} + {c^2}} } \right|}}$
Inside the modulus, we only calculate the numerical terms.
Now apply the values in the formula, we get,
The unit vector is $= \dfrac{{\left( { - 4\overrightarrow i + \overrightarrow j + 2\overrightarrow k } \right)}}{{\left| {\sqrt {\left( { - 4\overrightarrow i + \overrightarrow j + 2\overrightarrow k } \right)} } \right|}}$
Now simplify this term,
$= \dfrac{{\left( { - 4\overrightarrow i + \overrightarrow j + 2\overrightarrow k } \right)}}{{\left| {\sqrt {\left( { - {4^2} + {1^2} + {2^2}} \right)} } \right|}}$
Now we get, $= \dfrac{{\left( { - 4\overrightarrow i + \overrightarrow j + 2\overrightarrow k } \right)}}{{\left| {\sqrt {\left( {16 + 1 + 4} \right)} } \right|}}$
Hence we get,
Unit vector $= \dfrac{{\left( { - 4\overrightarrow i + \overrightarrow j + 2\overrightarrow k } \right)}}{{\sqrt {21} }}$
Length of the vector is $\sqrt 7$ perpendicular to $\overrightarrow A$and $\overrightarrow B$is,
$\dfrac{{\left( { - 4\overrightarrow i + \overrightarrow j + 2\overrightarrow k } \right)}}{{\sqrt {21} }} \times \sqrt 7$
$= \dfrac{{\left( { - 4\overrightarrow i + \overrightarrow j + 2\overrightarrow k } \right)}}{{\sqrt 7 \times \sqrt 3 }} \times \sqrt 7$
Now cancel the common term, we get
$= \dfrac{{\left( { - 4\overrightarrow i + \overrightarrow j + 2\overrightarrow k } \right)}}{{\sqrt 3 }}$
We can write this as, $= \dfrac{1}{{\sqrt 3 }}\left( { - 4\overrightarrow i + \overrightarrow j + 2\overrightarrow k } \right)$
Hene the answer for the given question is $= \dfrac{1}{{\sqrt 3 }}\left( { - 4\overrightarrow i + \overrightarrow j + 2\overrightarrow k } \right)$
Therefore the option C is a solution for the given question.
The explanation for Option A:
In this problem, we proved that the $= \dfrac{1}{{\sqrt 3 }}\left( { - 4\overrightarrow i + \overrightarrow j + 2\overrightarrow k } \right)$is the solution for the question. But in option A some values are changes.
Therefore option A is not an answer.
The explanation for option B:
In the above solution, we solved that option C is the solution for the given question.
Therefore option B is not the answer.
The explanation for option D:
From the above solution, we proved that the option C $= \dfrac{1}{{\sqrt 3 }}\left( { - 4\overrightarrow i + \overrightarrow j + 2\overrightarrow k } \right)$ is the solution for the given question,
Therefore option D is not an answer.

Option C is the correct answer.

Note:
In mathematics, Vector multiplication is defined as one of the techniques for the multiplication of two or more vectors. And the vector length is defined as the square root of the sum of vertical and horizontal terms. If a and b are zero there is no need for a vector length formula.