Question

Let $\overrightarrow a = \hat i + \hat j + \hat k,\overrightarrow b = \hat i - \hat j + \hat k$ and $\overrightarrow c = \hat i - \hat j - \hat k$ be three vectors. A vector $\overrightarrow v$ in the plane of $\overrightarrow a$ and $\overrightarrow b$ whose projection on $\overrightarrow c$ is $\dfrac{1}{{\sqrt 3 }}$ is given by:
A. $\hat i - 3\hat j + 3\hat k$
B. $- 3\hat i - 3\hat j - \hat k$
C. $3\hat i - \hat j + 3\hat k$
D. $\hat i + 3\hat j - 3\hat k$

Answer 1

Here we write the vector $\overrightarrow v$ in form of $\overrightarrow a$ and $\overrightarrow b$ using the concept of a vector in a plane made by other two vectors. Then we use the formula for projection of a vector on another vector and solve.

• A vector $\overrightarrow a$ in the plane of $\overrightarrow b$ and $\overrightarrow c$ is given by $\overrightarrow a = \overrightarrow b + \lambda \overrightarrow c$, where $\lambda$ is any constant value.
• Projection of a vector $\overrightarrow a$ on $\overrightarrow b$ is given by $\dfrac{{\overrightarrow a \bullet \overrightarrow b }}{{\left| {\overrightarrow b } \right|}}$, where the numerator has dot product of the vectors and the denominator has magnitude of a vector.
• For a vector $\overrightarrow b = x\hat i + y\hat j + z\hat k$ magnitude is calculated as $\left| {\overrightarrow b } \right| = \sqrt {{x^2} + {y^2} + {z^2}}$.
• For any two vectors $\overrightarrow a = x\hat i + y\hat j + z\hat k$ and $\overrightarrow b = x'\hat i + y'\hat j + z'\hat k$, the dot product is calculated as $\overrightarrow a \bullet \overrightarrow b = x.x' + y.y' + z.z'$

Complete step-by-step answer:
We have three vectors $\overrightarrow a = \hat i + \hat j + \hat k,\overrightarrow b = \hat i - \hat j + \hat k$ and $\overrightarrow c = \hat i - \hat j - \hat k$
Now we know that a new vector $\overrightarrow v$ is in the plane of $\overrightarrow a$ and $\overrightarrow b$
From the formula of a vector in the plane of other two vectors we can write
$\overrightarrow v = \overrightarrow a + \lambda \overrightarrow b$
Now we substitute the values of vectors $\overrightarrow a = \hat i + \hat j + \hat k,\overrightarrow b = \hat i - \hat j + \hat k$
$\Rightarrow \overrightarrow v = \hat i + \hat j + \hat k + \lambda (\hat i - \hat j + \hat k)$
Multiply the value outside the bracket to all terms in the bracket
$\Rightarrow \overrightarrow v = \hat i + \hat j + \hat k + \lambda \hat i - \lambda \hat j + \lambda \hat k$
Group together the terms of each direction.
$\Rightarrow \overrightarrow v = (\hat i + \lambda \hat i) + (\hat j - \lambda \hat j) + (\hat k + \lambda \hat k)$
By taking $\hat i,\hat j,\hat k$ common write the terms of vectors.
$\Rightarrow \overrightarrow v = (1 + \lambda )\hat i + (1 - \lambda )\hat j + (1 + \lambda )\hat k$ … (1)
Now we know that projection of $\overrightarrow v$ on $\overrightarrow c$ is $\dfrac{1}{{\sqrt 3 }}$
Using the formula of projection of a vector $\overrightarrow a$ on $\overrightarrow b$ is given by $\dfrac{{\overrightarrow a \bullet \overrightarrow b }}{{\left| {\overrightarrow b } \right|}}$
We can write
Projection $= \dfrac{{\overrightarrow v \bullet \overrightarrow c }}{{\left| {\overrightarrow c } \right|}}$ … (2)
First we find the dot product in the numerator
$\overrightarrow v \bullet \overrightarrow c = [(1 + \lambda )\hat i + (1 - \lambda )\hat j + (1 + \lambda )\hat k] \bullet [\hat i - \hat j - \hat k]$
We know for two vectors $\overrightarrow a = x\hat i + y\hat j + z\hat k$and $\overrightarrow b = x'\hat i + y'\hat j + z'\hat k$, the dot product is calculated as $\overrightarrow a \bullet \overrightarrow b = x.x' + y.y' + z.z'$.
$\Rightarrow \overrightarrow v \bullet \overrightarrow c = (1 + \lambda ).(1) + (1 - \lambda ).( - 1) + (1 + \lambda ).( - 1)$
Open the brackets and multiply the values
$\Rightarrow \overrightarrow v \bullet \overrightarrow c = 1 + \lambda - 1 + \lambda - 1 - \lambda$
Cancel all the possible terms having opposite signs but equal magnitudes
$\Rightarrow \overrightarrow v \bullet \overrightarrow c = \lambda - 1$
Now we calculate the magnitude of vector $\overrightarrow c$
We know $\overrightarrow c = \hat i - \hat j - \hat k$
We know a vector $\overrightarrow b = x\hat i + y\hat j + z\hat k$ has magnitude $\left| {\overrightarrow b } \right| = \sqrt {{x^2} + {y^2} + {z^2}}$
$\Rightarrow \left| {\overrightarrow c } \right| = \sqrt {{{(1)}^2} + {{( - 1)}^2} + {{( - 1)}^2}}$
Square the terms

$$\Rightarrow \left| {\overrightarrow c } \right| = \sqrt {1 + 1 + 1} \\\ \Rightarrow \left| {\overrightarrow c } \right| = \sqrt 3 \\\$$

Now we substitute the values of numerator and denominator in equation (2), and put the value of projection as $\dfrac{1}{{\sqrt 3 }}$
$\Rightarrow \dfrac{1}{{\sqrt 3 }} = \dfrac{{\lambda - 1}}{{\sqrt 3 }}$
Since both sides of the equation have the same denominator in the fraction, therefore we can cancel out the denominator.
$\Rightarrow \lambda - 1 = 1$
Shift the constant values on one side of the equation.

$$\Rightarrow \lambda = 1 + 1 \\\ \Rightarrow \lambda = 2 \\\$$

Now we substitute the value of $\lambda = 2$ in the equation (1)

$$\Rightarrow \overrightarrow v = (1 + 2)\hat i + (1 - 2)\hat j + (1 + 2)\hat k \\\ \Rightarrow \overrightarrow v = 3\hat i - \hat j + 3\hat k \\\$$

So, the correct answer is “Option C”.

Note: Students make mistake of considering dot product as normal multiplication and they multiply all the values from first bracket to all values in second bracket one by one which is wrong because when we take dot product of two vectors, we only write the multiplication of numbers in same direction because $\hat i.\hat i = \hat j.\hat j = \hat k.\hat k = 1$, as multiplication of numbers in different direction will give us zero because $\hat i.\hat j = \hat j.\hat k = \hat k.\hat i = 0$.