Select CORRECT statement(s) for three vectors `veca=-3hati+2hatj-hatk, vecb=hati-3hatj+5hatk` and `vecc=2hati+hatj-4hatk`

Options

A. The above vectors acna form triangle
B. Component of ` veca` along `vec c` is 3
C. `vec a` makes angle `cos^(-1)sqrt((2)/(7))` with y-axis.
D. A vector having magnitude twice the vector `veca` and anti parallel to vector `vecb` is `sqrt((2)/(5))(-2hati+6hatj-10hatk)`

Answer

To solve the problem, we need to analyze the three vectors given and check the validity of the statements regarding them. The vectors are:

\[ \vec{a} = -3\hat{i} + 2\hat{j} - \hat{k} \]
\[ \vec{b} = \hat{i} - 3\hat{j} + 5\hat{k} \]
\[ \vec{c} = 2\hat{i} + \hat{j} - 4\hat{k} \]

### Step 1: Check if the vectors form a triangle

To determine if the vectors form a triangle, we can use the triangle law of vector addition. According to this law, three vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) will form a triangle if:

\[ \vec{a} + \vec{b} + \vec{c} = \vec{0} \]

First, we calculate \(\vec{a} + \vec{b} + \vec{c}\):

\[ \vec{a} + \vec{b} = (-3\hat{i} + 2\hat{j} - \hat{k}) + (\hat{i} - 3\hat{j} + 5\hat{k}) \]
\[ = (-3 + 1)\hat{i} + (2 - 3)\hat{j} + (-1 + 5)\hat{k} \]
\[ = -2\hat{i} - \hat{j} + 4\hat{k} \]

Now, adding \(\vec{c}\):

\[ \vec{a} + \vec{b} + \vec{c} = (-2\hat{i} - \hat{j} + 4\hat{k}) + (2\hat{i} + \hat{j} - 4\hat{k}) \]
\[ = (-2 + 2)\hat{i} + (-1 + 1)\hat{j} + (4 - 4)\hat{k} \]
\[ = 0\hat{i} + 0\hat{j} + 0\hat{k} = \vec{0} \]

Since \(\vec{a} + \vec{b} + \vec{c} = \vec{0}\), the vectors do form a triangle.

### Step 2: Calculate the component of \(\vec{a}\) along \(\vec{c}\)

The component of vector \(\vec{a}\) along vector \(\vec{c}\) can be calculated using the formula:

\[ \text{Component of } \vec{a} \text{ along } \vec{c} = \frac{\vec{a} \cdot \vec{c}}{|\vec{c}|} \]

First, we calculate the dot product \(\vec{a} \cdot \vec{c}\):

\[ \vec{a} \cdot \vec{c} = (-3)(2) + (2)(1) + (-1)(-4) = -6 + 2 + 4 = 0 \]

Now, we calculate the magnitude of \(\vec{c}\):

\[ |\vec{c}| = \sqrt{(2)^2 + (1)^2 + (-4)^2} = \sqrt{4 + 1 + 16} = \sqrt{21} \]

Thus, the component of \(\vec{a}\) along \(\vec{c}\) is:

\[ \text{Component of } \vec{a} \text{ along } \vec{c} = \frac{0}{\sqrt{21}} = 0 \]

### Step 3: Calculate the angle between \(\vec{a}\) and the y-axis

To find the angle between \(\vec{a}\) and the y-axis, we use the cosine of the angle:

\[ \cos \theta = \frac{y \text{-component of } \vec{a}}{|\vec{a}|} \]

The y-component of \(\vec{a}\) is 2, and we calculate the magnitude of \(\vec{a}\):

\[ |\vec{a}| = \sqrt{(-3)^2 + (2)^2 + (-1)^2} = \sqrt{9 + 4 + 1} = \sqrt{14} \]

Thus,

\[ \cos \theta = \frac{2}{\sqrt{14}} = \frac{2}{\sqrt{2 \cdot 7}} = \frac{\sqrt{2}}{\sqrt{7}} \]

To find \(\theta\):

\[ \theta = \cos^{-1}\left(\frac{\sqrt{2}}{\sqrt{7}}\right) \]

### Conclusion

1. The vectors \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) form a triangle.
2. The component of \(\vec{a}\) along \(\vec{c}\) is 0.
3. The angle between \(\vec{a}\) and the y-axis is given by \(\theta = \cos^{-1}\left(\frac{\sqrt{2}}{\sqrt{7}}\right)\).