Understanding Vector Decomposition

This problem involves the concept of vector decomposition, specifically breaking down a vector \({\rm{\vec b}}\) into two components: one that is parallel to a given vector \({\rm{\vec a}}\) and another that is perpendicular to \({\rm{\vec a}}\). We are given two vectors, \({\rm{\vec a}} = {\rm{\hat i}} + {\rm{\hat j}}\) and \({\rm{\vec b}} = 3{\rm{\hat i}} + 4{\rm{\hat k}}\). We are told that \({\rm{\vec b}} = {\rm{\vec c}} + {\rm{\vec d}}\), where \({\rm{\vec c}}\) is parallel to \({\rm{\vec a}}\) and \({\rm{\vec d}}\) is perpendicular to \({\rm{\vec a}}\). Our goal is to find the vector \({\rm{\vec c}}\).

Setting up the Equations

Since \({\rm{\vec c}}\) is parallel to \({\rm{\vec a}}\), it can be written as a scalar multiple of \({\rm{\vec a}}\). Let's say \({\rm{\vec c}} = k{\rm{\vec a}}\) for some scalar \(k\). The given equation is \({\rm{\vec b}} = {\rm{\vec c}} + {\rm{\vec d}}\). Substituting \({\rm{\vec c}} = k{\rm{\vec a}}\), we get:

\({\rm{\vec b}} = k{\rm{\vec a}} + {\rm{\vec d}}\)

We also know that \({\rm{\vec d}}\) is perpendicular to \({\rm{\vec a}}\). This means their dot product is zero:

\({\rm{\vec d}} \cdot {\rm{\vec a}} = 0\)

Finding the Scalar Component (k)

To find the scalar \(k\), we can take the dot product of the equation \({\rm{\vec b}} = k{\rm{\vec a}} + {\rm{\vec d}}\) with \({\rm{\vec a}}\):

\({\rm{\vec b}} \cdot {\rm{\vec a}} = (k{\rm{\vec a}} + {\rm{\vec d}}) \cdot {\rm{\vec a}}\)

Using the properties of the dot product, we can distribute it:

\({\rm{\vec b}} \cdot {\rm{\vec a}} = k({\rm{\vec a}} \cdot {\rm{\vec a}}) + ({\rm{\vec d}} \cdot {\rm{\vec a}})\)

Since \({\rm{\vec d}} \cdot {\rm{\vec a}} = 0\) (because \({\rm{\vec d}}\) is perpendicular to \({\rm{\vec a}}\)), the equation simplifies to:

\({\rm{\vec b}} \cdot {\rm{\vec a}} = k({\rm{\vec a}} \cdot {\rm{\vec a}})\)

The dot product \({\rm{\vec a}} \cdot {\rm{\vec a}}\) is equal to the square of the magnitude of \({\rm{\vec a}}\), i.e., \(||\vec a||^2\). So, we have:

\({\rm{\vec b}} \cdot {\rm{\vec a}} = k ||\vec a||^2\)

We can solve for \(k\):

\(k = \frac{{{\rm{\vec b}} \cdot {\rm{\vec a}}}}{{||{\rm{\vec a}}||^2}}\)

This scalar \(k\) is also known as the scalar projection of \({\rm{\vec b}}\) onto \({\rm{\vec a}}\).

Calculating the Dot Product and Magnitude Squared

Let's calculate \({\rm{\vec b}} \cdot {\rm{\vec a}}\) and \(||{\rm{\vec a}}||^2\).

Given vectors:

• \({\rm{\vec a}} = {\rm{\hat i}} + {\rm{\hat j}} = 1{\rm{\hat i}} + 1{\rm{\hat j}} + 0{\rm{\hat k}}\)
• \({\rm{\vec b}} = 3{\rm{\hat i}} + 4{\rm{\hat k}} = 3{\rm{\hat i}} + 0{\rm{\hat j}} + 4{\rm{\hat k}}\)

The dot product \({\rm{\vec b}} \cdot {\rm{\vec a}}\) is:

\({\rm{\vec b}} \cdot {\rm{\vec a}} = (3)(1) + (0)(1) + (4)(0) = 3 + 0 + 0 = 3\)

The magnitude squared of \({\rm{\vec a}}\), \(||{\rm{\vec a}}||^2 = {\rm{\vec a}} \cdot {\rm{\vec a}}\), is:

\({\rm{\vec a}} \cdot {\rm{\vec a}} = (1)(1) + (1)(1) + (0)(0) = 1 + 1 + 0 = 2\)

Finding Vector c

Now we can find the scalar \(k\):

\(k = \frac{{{\rm{\vec b}} \cdot {\rm{\vec a}}}}{{||{\rm{\vec a}}||^2}} = \frac{3}{2}\)

Finally, we can find the vector \({\rm{\vec c}}\) using \({\rm{\vec c}} = k{\rm{\vec a}}\):

\({\rm{\vec c}} = \frac{3}{2}({\rm{\hat i}} + {\rm{\hat j}})\)

Verifying the Result

Let's check if this value of \({\rm{\vec c}}\) is among the given options.

Option 1: \(\frac{{3\left( {{\rm{\hat i}} + {\rm{\hat j}}} \right)}}{2}\)

Option 2: \(\frac{{2\left( {{\rm{\hat i}} + {\rm{\hat j}}} \right)}}{3}\)

Option 3: \(\frac{{\left( {{\rm{\hat i}} + {\rm{\hat j}}} \right)}}{2}\)

Option 4: \(\frac{{\left( {{\rm{\hat i}} + {\rm{\hat j}}} \right)}}{3}\)

Our calculated vector \({\rm{\vec c}} = \frac{3}{2}({\rm{\hat i}} + {\rm{\hat j}})\) matches Option 1.

Revision Table: Key Concepts

Additional Information: Vector Projection

The vector \({\rm{\vec c}}\) that is parallel to \({\rm{\vec a}}\) and part of the decomposition of \({\rm{\vec b}}\) is precisely the vector projection of \({\rm{\vec b}}\) onto \({\rm{\vec a}}\). The formula for the vector projection of \({\rm{\vec b}}\) onto \({\rm{\vec a}}\) is given by:

\(\text{proj}_{{\rm{\vec a}}} {\rm{\vec b}} = \frac{{{\rm{\vec b}} \cdot {\rm{\vec a}}}}{{||{\rm{\vec a}}||^2}} {\rm{\vec a}}\)

Comparing this formula to our result \({\rm{\vec c}} = \frac{{{\rm{\vec b}} \cdot {\rm{\vec a}}}}{{||{\rm{\vec a}}||^2}} {\rm{\vec a}}\), we see that \({\rm{\vec c}}\) is indeed the vector projection of \({\rm{\vec b}}\) onto \({\rm{\vec a}}\). This confirms our approach and result.

The other component, \({\rm{\vec d}}\), which is perpendicular to \({\rm{\vec a}}\), is given by \({\rm{\vec d}} = {\rm{\vec b}} - {\rm{\vec c}}\). This component is sometimes called the vector rejection of \({\rm{\vec b}}\) from \({\rm{\vec a}}\).