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JEE Main 20248 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
8 April
Shift
Morning Shift
Subject
Mathematics
Chapter
Vector Algebra

JEE Main 2024 · 8 April · Morning Shift · Mathematics · Vector Algebra

Let a =9 i -13 j +25 k , b =3 i +7 j -13 k and c =17 i -2 j + k be...

Let a⃗=9i^−13j^+25k^,b⃗=3i^+7j^−13k^\vec{a}=9 \hat{i}-13 \hat{j}+25 \hat{k}, \vec{b}=3 \hat{i}+7 \hat{j}-13 \hat{k} and c⃗=17i^−2j^+k^\vec{c}=17 \hat{i}-2 \hat{j}+\hat{k} be three given vectors. If r⃗\vec{r} is a vector such that r⃗×a⃗=(b⃗+c⃗)×a⃗\vec{r} \times \vec{a}=(\vec{b}+\vec{c}) \times \vec{a} and r⃗⋅(b⃗−c⃗)=0\vec{r} \cdot(\vec{b}-\vec{c})=0, then ∣593r⃗+67a⃗∣2(593)2\frac{|593 \vec{r}+67 \vec{a}|^2}{(593)^2} is equal to___________

Correct answer

0

Step-by-step solution

a→=9i^−13j^+25k^b→=3i^+7j^−13k^c→=17i^−2j^+k^b→+c→=20i^+5j^−12k^b→−c→=−14i^+9j^−14k^(r→−(b→+c→))×a=0r−(b→+c→)=λa→r→=λa→+b→+c→ But r→⋅(b→−c→)=0⇒(λa→+b→+c→)⋅(b→−c→)=0⇒λa→⋅b→+b→⋅b→+c→⋅b→−λa→⋅c→−b→⋅c→−c→⋅c→=0λ=c→⋅c→−b→⋅b→a→⋅b→−a→⋅c→=294−227−389=204=−67593∴r→=b→+c→−67593a→⇒593r→+67a→=593(b→+c→)⇒∣b→+c→∣2=569\begin{aligned} & \overrightarrow{\mathrm{a}}=9 \hat{\mathrm{i}}-13 \hat{\mathrm{j}}+25 \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{b}}=3 \hat{\mathrm{i}}+7 \hat{\mathrm{j}}-13 \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{c}}=17 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=20 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}-12 \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{c}}=-14 \hat{\mathrm{i}}+9 \hat{\mathrm{j}}-14 \hat{\mathrm{k}} \\ & (\overrightarrow{\mathrm{r}}-(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}))_{\times \mathrm{a}}=0 \\ & \mathrm{r}-(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})=\lambda \overrightarrow{\mathrm{a}} \\ & \overrightarrow{\mathrm{r}}=\lambda \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}} \\ & \text{ But } \overrightarrow{\mathrm{r}} \cdot(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{c}})=0 \\ & \Rightarrow(\lambda \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}) \cdot(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{c}})=0 \\ & \Rightarrow \lambda \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{b}}-\lambda \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{c}}=0 \\ & \lambda=\frac{\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{c}}-\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{b}}}{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{c}}}=\frac{294-227}{-389=204}=\frac{-67}{593} \\ & \therefore \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}-\frac{67}{593} \overrightarrow{\mathrm{a}} \\ & \Rightarrow 593 \overrightarrow{\mathrm{r}}+67 \overrightarrow{\mathrm{a}}=593(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}) \\ & \Rightarrow|\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}|^2=569\end{aligned}
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