Question

Consider three mutually perpendicular vectors $\vec{a}=2\hat{i}+3\hat{j}+6\hat{k}$, $\vec{b}=3\hat{i}-6\hat{j}+2\hat{k}$ and $\vec{c}=6\hat{i}+2\hat{j}-3\hat{k}$ and a vector $\vec{r}$ such that $|\vec{r}\times\hat{i}|^2+|\vec{r}\times\hat{j}|^2+|\vec{r}\times\hat{k}|^2=8$, $|\vec{r}\times\vec{a}|^2+|\vec{r}\times\vec{b}|^2+|\vec{r}\times\vec{c}|^2=k_1$ and $|\vec{r}\times(\vec{a}\times\vec{b})|^2+|\vec{r}\times(\vec{b}\times\vec{c})|^2+|\vec{r}\times(\vec{c}\times\vec{a})|^2=k_2$ Then Match the engries of List-l with the correct entries of List-II

• A. (P) Digit at unit place of the number $k_1$ is
• B. (Q) Digit at tens place of the number $k_1$ is
• C. (R) Digit at unit place of the number $k_2$ is
• D. (S) Digit at tens place of the number $k_2$ is
• E. (P) $\rightarrow$ (2), (Q) $\rightarrow$ (5), (R) $\rightarrow$ (4), (S) $\rightarrow$ (1)

Answer 1

Answer: (E)

(P) $\rightarrow$ (2), (Q) $\rightarrow$ (5), (R) $\rightarrow$ (4), (S) $\rightarrow$ (1)

Answer 2

1. Calculate $|\vec{r}|^2$: The condition $|\vec{r}\times\hat{i}|^2+|\vec{r}\times\hat{j}|^2+|\vec{r}\times\hat{k}|^2=8$ simplifies to $2|\vec{r}|^2 = 8$, which means $|\vec{r}|^2 = 4$.

2. Calculate $k_1$: $k_1 = |\vec{r}\times\vec{a}|^2+|\vec{r}\times\vec{b}|^2+|\vec{r}\times\vec{c}|^2$ Using the identity $|\vec{u}\times\vec{v}|^2 = |\vec{u}|^2|\vec{v}|^2 - (\vec{u}\cdot\vec{v})^2$: $k_1 = \sum_{\text{cyc}} (|\vec{r}|^2|\vec{a}|^2 - (\vec{r}\cdot\vec{a})^2)$ $k_1 = |\vec{r}|^2 (|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2) - \sum_{\text{cyc}} (\vec{r}\cdot\vec{a})^2$ Calculate magnitudes: $|\vec{a}| = \sqrt{2^2+3^2+6^2} = \sqrt{4+9+36} = \sqrt{49} = 7$. $|\vec{b}| = \sqrt{3^2+(-6)^2+2^2} = \sqrt{9+36+4} = \sqrt{49} = 7$. $|\vec{c}| = \sqrt{6^2+2^2+(-3)^2} = \sqrt{36+4+9} = \sqrt{49} = 7$. So, $|\vec{a}|^2=|\vec{b}|^2=|\vec{c}|^2=49$. Let $\vec{r} = x\hat{i}+y\hat{j}+z\hat{k}$. Then $|\vec{r}|^2 = x^2+y^2+z^2 = 4$. The sum of dot products squared is $\sum_{\text{cyc}} (\vec{r}\cdot\vec{a})^2 = (\vec{r}\cdot\vec{a})^2 + (\vec{r}\cdot\vec{b})^2 + (\vec{r}\cdot\vec{c})^2$. Since $\vec{a}, \vec{b}, \vec{c}$ are mutually perpendicular and have the same magnitude $7$, they form a scaled orthogonal basis. Let $\vec{r} = r_a \frac{\vec{a}}{|\vec{a}|} + r_b \frac{\vec{b}}{|\vec{b}|} + r_c \frac{\vec{c}}{|\vec{c}|}$. Then $\vec{r} \cdot \vec{a} = r_a |\vec{a}| = 7r_a$. $\vec{r} \cdot \vec{b} = r_b |\vec{b}| = 7r_b$. $\vec{r} \cdot \vec{c} = r_c |\vec{c}| = 7r_c$. $|\vec{r}|^2 = r_a^2 + r_b^2 + r_c^2 = 4$. $\sum_{\text{cyc}} (\vec{r}\cdot\vec{a})^2 = (7r_a)^2 + (7r_b)^2 + (7r_c)^2 = 49(r_a^2+r_b^2+r_c^2) = 49|\vec{r}|^2$. Substituting back into $k_1$: $k_1 = |\vec{r}|^2 (49+49+49) - 49|\vec{r}|^2$ $k_1 = |\vec{r}|^2 (3 \times 49) - 49 |\vec{r}|^2 = 2 \times 49 |\vec{r}|^2$ $k_1 = 2 \times 49 \times 4 = 392$. The digit at the unit place of $k_1$ is 2. (P) $\rightarrow$ (2). The digit at the tens place of $k_1$ is 9. (Q) $\rightarrow$ (5).

3. Calculate $k_2$: $k_2 = |\vec{r}\times(\vec{a}\times\vec{b})|^2+|\vec{r}\times(\vec{b}\times\vec{c})|^2+|\vec{r}\times(\vec{c}\times\vec{a})|^2$. Since $\vec{a}$, $\vec{b}$, $\vec{c}$ are mutually perpendicular, $\vec{a}\times\vec{b}$ is parallel to $\vec{c}$, $\vec{b}\times\vec{c}$ is parallel to $\vec{a}$, and $\vec{c}\times\vec{a}$ is parallel to $\vec{b}$. $\vec{a}\times\vec{b} = (|\vec{a}||\vec{b}|\sin 90^\circ) \hat{c} = (7 \times 7) \hat{c} = 49\hat{c}$. Similarly, $\vec{b}\times\vec{c} = 49\hat{a}$ and $\vec{c}\times\vec{a} = 49\hat{b}$. $k_2 = |\vec{r}\times(49\hat{c})|^2 + |\vec{r}\times(49\hat{a})|^2 + |\vec{r}\times(49\hat{b})|^2$ $k_2 = 49^2 (|\vec{r}\times\hat{a}|^2 + |\vec{r}\times\hat{b}|^2 + |\vec{r}\times\hat{c}|^2)$ We know that $|\vec{r}\times\hat{a}|^2 + |\vec{r}\times\hat{b}|^2 + |\vec{r}\times\hat{c}|^2 = 2|\vec{r}|^2$. $k_2 = 49^2 (2|\vec{r}|^2) = 2401 \times (2 \times 4) = 2401 \times 8 = 19208$. The digit at the unit place of $k_2$ is 8. (R) $\rightarrow$ (4). The digit at the tens place of $k_2$ is 0. (S) $\rightarrow$ (1).

4. Matching the lists: (P) Digit at unit place of $k_1$ is 2. $\rightarrow$ (2) (Q) Digit at tens place of $k_1$ is 9. $\rightarrow$ (5) (R) Digit at unit place of $k_2$ is 8. $\rightarrow$ (4) (S) Digit at tens place of $k_2$ is 0. $\rightarrow$ (1) The correct matching is (P) $\rightarrow$ (2), (Q) $\rightarrow$ (5), (R) $\rightarrow$ (4), (S) $\rightarrow$ (1).