Let \veca=2 \hati+\hatj-\hatk, \vecb=((\veca \times(\hati+\hatj)) \times \hati) \times \hati . Then the square of the projection of \veca on \vecb is:

Vector Algebra PYQ — JEE Main 2024 · 6 Apr · Shift 2 · Q37 with Solution

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Correct answer: C

Given vectors

$\vec a = 2\hat i + \hat j - \hat k$

and

$\vec b = \big((\vec a \times (\hat i + \hat j)) \times \hat i\big) \times \hat i.$

We need the square of the projection of $\vec a$ on $\vec b$ .

The scalar projection of $\vec a$ on $\vec b$ is

$\text{proj}_{\vec b}(\vec a) = \frac{\vec a \cdot \vec b}{|\vec b|}.$

So its square is

$\left(\frac{\vec a \cdot \vec b}{|\vec b|}\right)^2.$

Compute $\vec a \times (\hat i + \hat j)$

First,

$\hat i + \hat j = (1,1,0), \qquad \vec a = (2,1,-1).$

So,

= \begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & 1 & -1 \\ 1 & 1 & 0 \end{vmatrix}.$$ Expanding: $$= \hat i(1\cdot 0 - (-1)\cdot 1) - \hat j(2\cdot 0 - (-1)\cdot 1) + \hat k(2\cdot 1 - 1\cdot 1)$$ $$= \hat i(1) - \hat j(1) + \hat k(1)$$ $$= \hat i - \hat j + \hat k.$$ Let $$\vec c = \vec a \times (\hat i + \hat j) = \hat i - \hat j + \hat k.$$ --- 3. **Compute** $(\vec c \times \hat i)$ $$\vec c = (1,-1,1), \qquad \hat i = (1,0,0).$$ Then $$\vec c \times \hat i = \begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & -1 & 1 \\ 1 & 0 & 0 \end{vmatrix}.$$ Expanding: $$= \hat i((-1)\cdot 0 - 1\cdot 0) - \hat j(1\cdot 0 - 1\cdot 1) + \hat k(1\cdot 0 - (-1)\cdot 1)$$ $$= 0\hat i - \hat j(-1) + \hat k(1)$$ $$= \hat j + \hat k.$$ So, $$\vec d = (\vec c \times \hat i) = \hat j + \hat k.$$ --- 4. **Compute** $\vec b = \vec d \times \hat i$ $$\vec d = (0,1,1).$$ Thus, $$\vec b = \vec d \times \hat i = \begin{vmatrix} \hat i & \hat j & \hat k \\ 0 & 1 & 1 \\ 1 & 0 & 0 \end{vmatrix}.$$ Expanding: $$= \hat i(1\cdot 0 - 1\cdot 0) - \hat j(0\cdot 0 - 1\cdot 1) + \hat k(0\cdot 0 - 1\cdot 1)$$ $$= 0\hat i - \hat j(-1) + \hat k(-1)$$ $$= \hat j - \hat k.$$ So, $$\vec b = \hat j - \hat k = (0,1,-1).$$ --- 5. **Find the projection of** $\vec a$ **on** $\vec b$ Compute dot product: $$\vec a \cdot \vec b = (2,1,-1)\cdot(0,1,-1) = 0 + 1 + 1 = 2.$$ Magnitude of $\vec b$: $$|\vec b| = \sqrt{0^2 + 1^2 + (-1)^2} = \sqrt{2}.$$ Therefore scalar projection is $$\frac{\vec a \cdot \vec b}{|\vec b|} = \frac{2}{\sqrt{2}} = \sqrt{2}.$$ Hence the **square** of the projection is $$\left(\sqrt{2}\right)^2 = 2.$$ --- 6. **Check options** - A: $\frac{1}{3}$ - B: $\frac{1}{5}$ - C: $2$ - D: $\frac{2}{3}$ Thus the correct option is $$\boxed{\text{C }(2)}.$$ --- 7. **Comparison with stored answer** Stored correct answer: **C** Our derived answer: **C** So they agree.

More from Vector Algebra

View written solutionFree

Correct answer: C

Given vectors

$\vec a = 2\hat i + \hat j - \hat k$

and

$\vec b = \big((\vec a \times (\hat i + \hat j)) \times \hat i\big) \times \hat i.$

We need the square of the projection of $\vec a$ on $\vec b$ .

The scalar projection of $\vec a$ on $\vec b$ is

$\text{proj}_{\vec b}(\vec a) = \frac{\vec a \cdot \vec b}{|\vec b|}.$

So its square is

$\left(\frac{\vec a \cdot \vec b}{|\vec b|}\right)^2.$

Compute $\vec a \times (\hat i + \hat j)$

First,

$\hat i + \hat j = (1,1,0), \qquad \vec a = (2,1,-1).$

So,

= \begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & 1 & -1 \\ 1 & 1 & 0 \end{vmatrix}.$$ Expanding: $$= \hat i(1\cdot 0 - (-1)\cdot 1) - \hat j(2\cdot 0 - (-1)\cdot 1) + \hat k(2\cdot 1 - 1\cdot 1)$$ $$= \hat i(1) - \hat j(1) + \hat k(1)$$ $$= \hat i - \hat j + \hat k.$$ Let $$\vec c = \vec a \times (\hat i + \hat j) = \hat i - \hat j + \hat k.$$ --- 3. **Compute** $(\vec c \times \hat i)$ $$\vec c = (1,-1,1), \qquad \hat i = (1,0,0).$$ Then $$\vec c \times \hat i = \begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & -1 & 1 \\ 1 & 0 & 0 \end{vmatrix}.$$ Expanding: $$= \hat i((-1)\cdot 0 - 1\cdot 0) - \hat j(1\cdot 0 - 1\cdot 1) + \hat k(1\cdot 0 - (-1)\cdot 1)$$ $$= 0\hat i - \hat j(-1) + \hat k(1)$$ $$= \hat j + \hat k.$$ So, $$\vec d = (\vec c \times \hat i) = \hat j + \hat k.$$ --- 4. **Compute** $\vec b = \vec d \times \hat i$ $$\vec d = (0,1,1).$$ Thus, $$\vec b = \vec d \times \hat i = \begin{vmatrix} \hat i & \hat j & \hat k \\ 0 & 1 & 1 \\ 1 & 0 & 0 \end{vmatrix}.$$ Expanding: $$= \hat i(1\cdot 0 - 1\cdot 0) - \hat j(0\cdot 0 - 1\cdot 1) + \hat k(0\cdot 0 - 1\cdot 1)$$ $$= 0\hat i - \hat j(-1) + \hat k(-1)$$ $$= \hat j - \hat k.$$ So, $$\vec b = \hat j - \hat k = (0,1,-1).$$ --- 5. **Find the projection of** $\vec a$ **on** $\vec b$ Compute dot product: $$\vec a \cdot \vec b = (2,1,-1)\cdot(0,1,-1) = 0 + 1 + 1 = 2.$$ Magnitude of $\vec b$: $$|\vec b| = \sqrt{0^2 + 1^2 + (-1)^2} = \sqrt{2}.$$ Therefore scalar projection is $$\frac{\vec a \cdot \vec b}{|\vec b|} = \frac{2}{\sqrt{2}} = \sqrt{2}.$$ Hence the **square** of the projection is $$\left(\sqrt{2}\right)^2 = 2.$$ --- 6. **Check options** - A: $\frac{1}{3}$ - B: $\frac{1}{5}$ - C: $2$ - D: $\frac{2}{3}$ Thus the correct option is $$\boxed{\text{C }(2)}.$$ --- 7. **Comparison with stored answer** Stored correct answer: **C** Our derived answer: **C** So they agree.

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