Let \overrightarrow a be a vector which is perpendicular to the vector 3\widehat i + 1 \over 2\widehat j + 2\widehat k . If \overrightarrow a \times ( 2\widehat i + \widehat k ) = 2\widehat i - 13\widehat j - 4\widehat k , then the projection of the vector \overrightarrow a on the vector 2\widehat i + 2\widehat j + \widehat k is :

Vector Algebra PYQ — JEE Main 2022 · 28 Jun · Shift 2 · Q39 with Solution

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Correct answer: C

We are given two conditions.

Since $\vec a$ is perpendicular to $3\hat i + \frac12 \hat j + 2\hat k$ , their dot product is zero: $\vec a \cdot \left(3\hat i + \frac12 \hat j + 2\hat k\right)=0.$ So, $3x + \frac12 y + 2z = 0.$ Multiplying by $2$ : $6x + y + 4z = 0. \qquad (1)$

Now use the cross product condition: $\vec a \times (2\hat i + \hat k) = 2\hat i - 13\hat j - 4\hat k.$

Write $2\hat i + \hat k = (2,0,1)$ . Then

= \begin{vmatrix} \hat i & \hat j & \hat k\\ x & y & z\\ 2 & 0 & 1 \end{vmatrix}.$$ Expanding: $$\vec a \times (2,0,1) = \hat i(y\cdot 1 - z\cdot 0) - \hat j(x\cdot 1 - z\cdot 2) + \hat k(x\cdot 0 - y\cdot 2).$$ Thus, $$\vec a \times (2\hat i + \hat k)= y\hat i -(x-2z)\hat j -2y\hat k.$$ Comparing with $$2\hat i -13\hat j -4\hat k,$$ we get: $$y=2,$$ $$-(x-2z)=-13 \Rightarrow x-2z=13,$$ which is $$x=13+2z. \qquad (2)$$ Also, $$-2y=-4 \Rightarrow y=2,$$ consistent. --- 4. Substitute $y=2$ and $x=13+2z$ into equation (1): $$6x + y + 4z = 0$$ $$6(13+2z)+2+4z=0$$ $$78+12z+2+4z=0$$ $$80+16z=0$$ $$z=-5.$$ Then from (2), $$x=13+2(-5)=3.$$ Hence, $$\vec a = 3\hat i + 2\hat j -5\hat k.$$ --- 5. We need the projection of $\vec a$ on $2\hat i + 2\hat j + \hat k$. Let $$\vec b = 2\hat i + 2\hat j + \hat k.$$ The scalar projection of $\vec a$ on $\vec b$ is $$\text{proj}_{\vec b}(\vec a) = \frac{\vec a \cdot \vec b}{|\vec b|}.$$ First compute the dot product: $$\vec a \cdot \vec b = (3)(2)+(2)(2)+(-5)(1)=6+4-5=5.$$ Now, $$|\vec b|=\sqrt{2^2+2^2+1^2} = \sqrt{9}=3.$$ Therefore, $$\text{projection} = \frac{5}{3}.$$ --- 6. So the correct option is $$\boxed{\frac53}$$ which is **Option C**.

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Correct answer: C

We are given two conditions.

Since $\vec a$ is perpendicular to $3\hat i + \frac12 \hat j + 2\hat k$ , their dot product is zero: $\vec a \cdot \left(3\hat i + \frac12 \hat j + 2\hat k\right)=0.$ So, $3x + \frac12 y + 2z = 0.$ Multiplying by $2$ : $6x + y + 4z = 0. \qquad (1)$

Now use the cross product condition: $\vec a \times (2\hat i + \hat k) = 2\hat i - 13\hat j - 4\hat k.$

Write $2\hat i + \hat k = (2,0,1)$ . Then

= \begin{vmatrix} \hat i & \hat j & \hat k\\ x & y & z\\ 2 & 0 & 1 \end{vmatrix}.$$ Expanding: $$\vec a \times (2,0,1) = \hat i(y\cdot 1 - z\cdot 0) - \hat j(x\cdot 1 - z\cdot 2) + \hat k(x\cdot 0 - y\cdot 2).$$ Thus, $$\vec a \times (2\hat i + \hat k)= y\hat i -(x-2z)\hat j -2y\hat k.$$ Comparing with $$2\hat i -13\hat j -4\hat k,$$ we get: $$y=2,$$ $$-(x-2z)=-13 \Rightarrow x-2z=13,$$ which is $$x=13+2z. \qquad (2)$$ Also, $$-2y=-4 \Rightarrow y=2,$$ consistent. --- 4. Substitute $y=2$ and $x=13+2z$ into equation (1): $$6x + y + 4z = 0$$ $$6(13+2z)+2+4z=0$$ $$78+12z+2+4z=0$$ $$80+16z=0$$ $$z=-5.$$ Then from (2), $$x=13+2(-5)=3.$$ Hence, $$\vec a = 3\hat i + 2\hat j -5\hat k.$$ --- 5. We need the projection of $\vec a$ on $2\hat i + 2\hat j + \hat k$. Let $$\vec b = 2\hat i + 2\hat j + \hat k.$$ The scalar projection of $\vec a$ on $\vec b$ is $$\text{proj}_{\vec b}(\vec a) = \frac{\vec a \cdot \vec b}{|\vec b|}.$$ First compute the dot product: $$\vec a \cdot \vec b = (3)(2)+(2)(2)+(-5)(1)=6+4-5=5.$$ Now, $$|\vec b|=\sqrt{2^2+2^2+1^2} = \sqrt{9}=3.$$ Therefore, $$\text{projection} = \frac{5}{3}.$$ --- 6. So the correct option is $$\boxed{\frac53}$$ which is **Option C**.