A vector A points vertically upward and B points towards north. The vector product \[\overrightarrow A \times \overrightarrow B\] is:
$(A)$ Null vector
$(B)$ Along west
$(C)$ Along east
$(D)$ Vertically downward
Answer
Verified
120k+ views
Hint: Draw a clear picture of the direction of the vectors along three axes. Note that, vertically upward means the direction along the normal to the plane of the paper in which we are drawing.
After getting the axis of the vectors represent the vectors with unit vectors related to those axes.
Here in the problem cross product is needed, so do the cross product of the vectors find the direction of the product based on the sign.
Formula used:
If $\widehat i,\widehat j,\widehat k$ are the unit vectors along $x,y,z$ axis respectively, then one of the relations between the unit vectors related to cross multiplications is :
$\widehat j \times \widehat k = \widehat i$
$ \Rightarrow \widehat k \times \widehat j = - \widehat i$
Complete step by step answer:
A vector $\overrightarrow A $ is along vertically upward and another vector $\overrightarrow B $ is along the north direction. If this is drawn in a plane of a paper it looks like:
$\overrightarrow A $is along the $z$ axis that is along the normal direction to the plane of the paper. $\overrightarrow B $ is along the $y$ axis that is along the north direction to the plane.
If $\widehat i,\widehat j,\widehat k$ are the unit vectors along the positive $x,y,z$ axis respectively, then
\[\overrightarrow A = A\widehat k\]
And, \[\overrightarrow B = B\widehat j\]
We know, one of the relations between the unit vectors related to cross multiplications is :
$\widehat j \times \widehat k = \widehat i$
$ \Rightarrow \widehat k \times \widehat j = - \widehat i$
Hence, $\overrightarrow A \times \overrightarrow B = A\widehat k \times B\widehat j = AB( - \widehat i)$
$\therefore \overrightarrow A \times \overrightarrow B = - AB\widehat i$
Since the unit vector is $\widehat i$ with a negative sign, the resultant vector will be along the negative $x$ axis. From the figure it can be seen that the negative $x$ axis is directed towards the west direction.
Hence the correct option is (B).
Note: $\widehat i,\widehat j,\widehat k$ are the unit vectors along the positive $x,y,z$ axis respectively. Each of them has a value $1$ .
Since $\overrightarrow A \times \overrightarrow A = 0$ ;
Hence, $\widehat i \times \widehat i = \widehat j \times \widehat j = \widehat k \times \widehat k = 0$ .
Again, the value of $\widehat i \times \widehat j = \left| {\widehat i \times \widehat j} \right| = \left( 1 \right).\left( 1 \right).\sin {90^ \circ } = 1$
The rule of right hand cork-screw, the direction of $\widehat i \times \widehat j$ is along the $z$ axis. Since $\widehat k$ is the unit vector along the $z$ axis, so $\widehat i \times \widehat j = \widehat k$
According to this rule, $\widehat i \times \widehat j = \widehat k$, $\widehat j \times \widehat k = \widehat i$ and, \[\widehat k \times \widehat i = \widehat j\]
Since, $\overrightarrow A \times \overrightarrow B = - \overrightarrow B \times \overrightarrow A $
Hence, $\widehat j \times \widehat i = - \widehat k$, $\widehat k \times \widehat j = - \widehat i$ and, \[\widehat i \times \widehat k = - \widehat j\]
After getting the axis of the vectors represent the vectors with unit vectors related to those axes.
Here in the problem cross product is needed, so do the cross product of the vectors find the direction of the product based on the sign.
Formula used:
If $\widehat i,\widehat j,\widehat k$ are the unit vectors along $x,y,z$ axis respectively, then one of the relations between the unit vectors related to cross multiplications is :
$\widehat j \times \widehat k = \widehat i$
$ \Rightarrow \widehat k \times \widehat j = - \widehat i$
Complete step by step answer:
A vector $\overrightarrow A $ is along vertically upward and another vector $\overrightarrow B $ is along the north direction. If this is drawn in a plane of a paper it looks like:
$\overrightarrow A $is along the $z$ axis that is along the normal direction to the plane of the paper. $\overrightarrow B $ is along the $y$ axis that is along the north direction to the plane.
If $\widehat i,\widehat j,\widehat k$ are the unit vectors along the positive $x,y,z$ axis respectively, then
\[\overrightarrow A = A\widehat k\]
And, \[\overrightarrow B = B\widehat j\]
We know, one of the relations between the unit vectors related to cross multiplications is :
$\widehat j \times \widehat k = \widehat i$
$ \Rightarrow \widehat k \times \widehat j = - \widehat i$
Hence, $\overrightarrow A \times \overrightarrow B = A\widehat k \times B\widehat j = AB( - \widehat i)$
$\therefore \overrightarrow A \times \overrightarrow B = - AB\widehat i$
Since the unit vector is $\widehat i$ with a negative sign, the resultant vector will be along the negative $x$ axis. From the figure it can be seen that the negative $x$ axis is directed towards the west direction.
Hence the correct option is (B).
Note: $\widehat i,\widehat j,\widehat k$ are the unit vectors along the positive $x,y,z$ axis respectively. Each of them has a value $1$ .
Since $\overrightarrow A \times \overrightarrow A = 0$ ;
Hence, $\widehat i \times \widehat i = \widehat j \times \widehat j = \widehat k \times \widehat k = 0$ .
Again, the value of $\widehat i \times \widehat j = \left| {\widehat i \times \widehat j} \right| = \left( 1 \right).\left( 1 \right).\sin {90^ \circ } = 1$
The rule of right hand cork-screw, the direction of $\widehat i \times \widehat j$ is along the $z$ axis. Since $\widehat k$ is the unit vector along the $z$ axis, so $\widehat i \times \widehat j = \widehat k$
According to this rule, $\widehat i \times \widehat j = \widehat k$, $\widehat j \times \widehat k = \widehat i$ and, \[\widehat k \times \widehat i = \widehat j\]
Since, $\overrightarrow A \times \overrightarrow B = - \overrightarrow B \times \overrightarrow A $
Hence, $\widehat j \times \widehat i = - \widehat k$, $\widehat k \times \widehat j = - \widehat i$ and, \[\widehat i \times \widehat k = - \widehat j\]
Recently Updated Pages
Difference Between Circuit Switching and Packet Switching
Difference Between Mass and Weight
JEE Main Participating Colleges 2024 - A Complete List of Top Colleges
JEE Main Maths Paper Pattern 2025 – Marking, Sections & Tips
Sign up for JEE Main 2025 Live Classes - Vedantu
JEE Main 2025 Helpline Numbers - Center Contact, Phone Number, Address
Trending doubts
JEE Mains 2025: Check Important Dates, Syllabus, Exam Pattern, Fee and Updates
JEE Main Login 2045: Step-by-Step Instructions and Details
Class 11 JEE Main Physics Mock Test 2025
JEE Main Chemistry Question Paper with Answer Keys and Solutions
JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking
JEE Main Chemistry Exam Pattern 2025
Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs
NCERT Solutions for Class 11 Physics Chapter 7 Gravitation
NCERT Solutions for Class 11 Physics Chapter 1 Units and Measurements
NCERT Solutions for Class 11 Physics Chapter 9 Mechanical Properties of Fluids
Units and Measurements Class 11 Notes - CBSE Physics Chapter 1
NCERT Solutions for Class 11 Physics Chapter 2 Motion In A Straight Line