Exams Joint Entrance Examination Mathematics Vector Algebra Question 102 Question 102 of 379 Medium NUMERICAL
Let a ⃗ = 9 i ^ − 13 j ^ + 25 k ^ , b ⃗ = 3 i ^ + 7 j ^ − 13 k ^ \vec{a}=9\hat{i}-13\hat{j}+25\hat{k},\vec{b}=3\hat{i}+7\hat{j}-13\hat{k} a = 9 i ^ − 13 j ^ + 25 k ^ , b = 3 i ^ + 7 j ^ − 13 k ^ and c ⃗ = 17 i ^ − 2 j ^ + k ^ \vec{c}=17\hat{i}-2\hat{j}+\hat{k} c = 17 i ^ − 2 j ^ + k ^ be three given vectors. If r ⃗ \vec{r} r is a vector such that r ⃗ × a ⃗ = ( b ⃗ + c ⃗ ) × a ⃗ \vec{r}\times \vec{a}=(\vec{b}+\vec{c})\times \vec{a} r × a = ( b + c ) × a and r ⃗ ⋅ ( b ⃗ − c ⃗ ) = 0 \vec{r}\cdot (\vec{b}-\vec{c})=0 r ⋅ ( b − c ) = 0 , then ∣ 593 r ⃗ + 67 a ⃗ ∣ 2 ( 593 ) 2 \frac{|593\vec{r}+67\vec{a}{|}^{2}}{(593{)}^{2}} ( 593 ) 2 ∣593 r + 67 a ∣ 2 is equal to \_\_\_\_\_\_\_\_\_\_.
Answer: 569
Solutions · 1 Post your solution Official solution
a ⃗ = 9 i ^ − 13 j ^ + 25 k ^ b ⃗ = 3 i ^ + 7 j ^ − 13 k ^ c ⃗ = 17 i ^ − 2 j ^ + k ^ r ⃗ × a ⃗ = ( b ⃗ + c ⃗ ) × a ⃗ ( r ⃗ − ( b ⃗ + c ⃗ ) ) × a ⃗ = 0 ⇒ r ⃗ = ( b ⃗ + c ⃗ ) + λ a ⃗ r ⃗ = ( 20 i ^ + 5 j ^ − 12 k ^ ) + λ ( 9 i ^ − 13 j ^ + 25 k ^ ) = ( 20 + 9 λ ) i ^ + ( 5 − 13 λ ) j ^ + ( 25 λ − 12 ) k ^ \begin{gathered} \vec{a}=9\hat{i}-13\hat{j}+25\hat{k} \\ \vec{b}=3\hat{i}+7\hat{j}-13\hat{k} \\ \vec{c}=17\hat{i}-2\hat{j}+\hat{k} \\ \vec{r}\times \vec{a}=(\vec{b}+\vec{c})\times \vec{a} \\ (\vec{r}-(\vec{b}+\vec{c}))\times \vec{a}=0 \\ \Rightarrow \vec{r}=(\vec{b}+\vec{c})+\lambda \vec{a} \\ \vec{r}=(20\hat{i}+5\hat{j}-12\hat{k})+\lambda (9\hat{i}-13\hat{j}+25\hat{k}) \\ =(20+9\lambda )\hat{i}+(5-13\lambda )\hat{j}+(25\lambda -12)\hat{k} \end{gathered} a = 9 i ^ − 13 j ^ + 25 k ^ b = 3 i ^ + 7 j ^ − 13 k ^ c = 17 i ^ − 2 j ^ + k ^ r × a = ( b + c ) × a ( r − ( b + c )) × a = 0 ⇒ r = ( b + c ) + λ a r = ( 20 i ^ + 5 j ^ − 12 k ^ ) + λ ( 9 i ^ − 13 j ^ + 25 k ^ ) = ( 20 + 9 λ ) i ^ + ( 5 − 13 λ ) j ^ + ( 25 λ − 12 ) k ^
Now r ⃗ ⋅ ( b ⃗ − c ⃗ ) = 0 \vec{r}\cdot (\vec{b}-\vec{c})=0 r ⋅ ( b − c ) = 0
r ⃗ ⋅ ( − 14 i ^ + 9 j ^ − 14 k ^ ) = 0 \vec{r}\cdot (-14\hat{i}+9\hat{j}-14\hat{k})=0 r ⋅ ( − 14 i ^ + 9 j ^ − 14 k ^ ) = 0
Now
− 14 ( 20 + 9 λ ) + 9 ( 5 − 13 λ ) − 14 ( 25 λ − 12 ) = 0 − 593 λ − 67 = 0 λ = − 67 593 ∴ r ⃗ = ( b ⃗ + c ⃗ ) − 67 593 a ⃗ ∣ 593 r ⃗ + 67 a ⃗ ∣ 2 ∣ 593 ∣ 2 = ∣ b ⃗ + c ⃗ ∣ 2 = ∣ 20 i ^ + 5 j ^ − 12 k ^ ∣ 2 = 569 \begin{gathered} -14(20+9\lambda )+9(5-13\lambda )-14(25\lambda -12)=0 \\ -593\lambda -67=0 \\ \lambda =-\frac{67}{593} \\ \therefore \vec{r}=(\vec{b}+\vec{c})-\frac{67}{593}\vec{a} \\ \frac{|593\vec{r}+67\vec{a}{|}^{2}}{|593{|}^{2}}=|\vec{b}+\vec{c}{|}^{2}=|20\hat{i}+5\hat{j}-12\hat{k}{|}^{2} \\ =569 \end{gathered} − 14 ( 20 + 9 λ ) + 9 ( 5 − 13 λ ) − 14 ( 25 λ − 12 ) = 0 − 593 λ − 67 = 0 λ = − 593 67 ∴ r = ( b + c ) − 593 67 a ∣593 ∣ 2 ∣593 r + 67 a ∣ 2 = ∣ b + c ∣ 2 = ∣20 i ^ + 5 j ^ − 12 k ^ ∣ 2 = 569 Answer the question to unlock