Question

Consider a vector $\vec A = 5\hat i + p\hat j + 4\sqrt 2 \hat k$ . Its modulus is 11. Find the value of $p$ .
A) +$8$
B) −$6$
C) ±$8$
D) ±$6$

Answer 1

Hint:Vectors have direction and magnitude. The modulus of a vector refers to the magnitude or length of the vector. Any vector can be resolved into three components along the x-axis, y-axis and z-axis in three dimensions. Now the modulus of the vector will be the square root of the sum of the squares of the three components. Here, $p$ represents the component of $\vec A$ along the y-axis.

Formulas used:
-The magnitude or modulus of a general vector $\vec B = {B_x}\hat i + {B_y}\hat j + {B_z}\hat k$ is given by, $\left| {\vec B} \right| = \sqrt {B_x^2 + B_y^2 + B_z^2} $ where ${B_x}$ , ${B_y}$ , ${B_z}$ are the components of $\vec B$ along the x, y, z axes respectively.

Complete step by step answer.
Step 1: List the data given in the question.
We have a vector $\vec A = 5\hat i + p\hat j + 4\sqrt 2 \hat k$ and its magnitude is given as 11.
The magnitude or modulus of a general vector $\vec B = {B_x}\hat i + {B_y}\hat j + {B_z}\hat k$ is given by, $\left| {\vec B} \right| = \sqrt {B_x^2 + B_y^2 + B_z^2} $ where ${B_x}$ , ${B_y}$ , ${B_z}$ are the components of $\vec B$ along the x, y, z axes respectively.
On comparing with the general form of the vector we can list the components of our vector along x, y and z axes.
Now we have, ${A_x} = 5$ , ${A_y} = p$ and ${A_z} = 4\sqrt 2 $ .
Step 2: Express the relation for the magnitude of the vector.
The magnitude or modulus of a general vector $\vec B = {B_x}\hat i + {B_y}\hat j + {B_z}\hat k$ is given by $\left| {\vec B} \right| = \sqrt {B_x^2 + B_y^2 + B_z^2} $ --------- (1)
where ${B_x}$ , ${B_y}$ , ${B_z}$ are the components of $\vec B$ along the x, y, z axes respectively.
From equation (1), the magnitude of $\vec A = 5\hat i + p\hat j + 4\sqrt 2 \hat k$ can be expressed as, $\left| {\vec A} \right| = \sqrt {A_x^2 + A_y^2 + A_z^2} $
Substituting the values for ${A_x} = 5$ , ${A_y} = p$ and ${A_z} = 4\sqrt 2 $ in the above equation we get, $\left| {\vec A} \right| = \sqrt {{5^2} + {p^2} + {{\left( {4\sqrt 2 } \right)}^2}} $
Simplifying we get, $\left| {\vec A} \right| = \sqrt {25 + {p^2} + 32} $ -------- (2)
Step 3: Find the value of $p$using equation (2)
Equation (2) gives us $\left| {\vec A} \right| = \sqrt {25 + {p^2} + 32} $ .
Also, it is given that the magnitude of the vector is 11 i.e., $\left| {\vec A} \right| = 11$.
We now substitute the value for $\left| {\vec A} \right| = 11$ in equation (2) to find the value of $p$ .
Thus we have, $\sqrt {25 + {p^2} + 32} = 11$
On squaring the above equation we get, $25 + {p^2} + 32 = 121$ .
Rearranging we get, ${p^2} = 121 - 32 - 25 = 64$
Take the square root to get the required value of $p$ .

Therefore, the value of $p = \pm 8$ .

Note: We can back substitute the obtained values of $p = \pm 8$ in equation (2) to check if the magnitude of our vector is 11 as given in the question.
Equation (2) gives us $\left| {\vec A} \right| = \sqrt {25 + {p^2} + 32} $
For $p = + 8$ we get, $\left| {\vec A} \right| = \sqrt {25 + {8^2} + 32} $ , so $\left| {\vec A} \right| = 11$
For $p = + 8$ we get, $\left| {\vec A} \right| = \sqrt {25 + {{\left( { - 8} \right)}^2} + 32} $ , so $\left| {\vec A} \right| = 11$