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}a=9i^−13j^+25k^,b=3i^+7j^−13k^ and c⃗=17i^−2j^+k^\vec{c}=17 \hat{i}-2 \hat{j}+\hat{k}c=17i^−2j^+k^ be three given vectors. If r⃗\vec{r}r is a vector such that r⃗×a⃗=(b⃗+c⃗)×a⃗\vec{r} \times \vec{a}=(\vec{b}+\vec{c}) \times \vec{a}r×a=(b+c)×a and r⃗⋅(b⃗−c⃗)=0\vec{r} \cdot(\vec{b}-\vec{c})=0r⋅(b−c)=0, then ∣593r⃗+67a⃗∣2(593)2\frac{|593 \vec{r}+67 \vec{a}|^2}{(593)^2}(593)2∣593r+67a∣2 is equal to___________
Held on 8 Apr 2024 · Verified 6 Jul 2026.
Sign in to track your attempts and accuracy.
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}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)=λar=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=a⋅b−a⋅cc⋅c−b⋅b=−389=204294−227=593−67∴r=b+c−59367a⇒593r+67a=593(b+c)⇒∣b+c∣2=569
Sign in to keep a private note on this question. Nothing you write is ever public.
Let the line $\mathrm{L}_{1}$ be parallel to the vector $-3 \hat{i}+2 \hat{j}+4 \hat{k}$ and pass through the point (2,6,7), and the line $\mathrm{L}_{2}$ be parallel to the vector $2 \hat{i}+\hat{j}+3 \hat{k}$ and pass through the point $(4,3,5)$. If the line $\mathrm{L}_{3}$ is parallel to the vector $-3 \hat{i}+5 \hat{j}+16 \hat{k}$ and intersects the lines $\mathrm{L}_{1}$ and $\mathrm{L}_{2}$ at the points C and D, respectively, then $|\overrightarrow{C D}|^{2}$ is equal to :
Let the image of the point $P(0, -5, 0)$ in the line $\dfrac{x-1}{2} = \dfrac{y}{1} = \dfrac{z+1}{-2}$ be the point $R$ and the image of the point $Q\left(0, \dfrac{-1}{2}, 0\right)$ in the line $\dfrac{x-1}{-1} = \dfrac{y+9}{4} = \dfrac{z+1}{1}$ be the point $S$. Then the square of the area of the parallelogram $PQRS$ is __________.
The square of the distance of the point $P(5, 6, 7)$ from the line $\dfrac{x-2}{2} = \dfrac{y-5}{3} = \dfrac{z-2}{4}$ is equal to:
Let $\overrightarrow{\mathrm{AB}}=2 \hat{i}+4 \hat{j}-5 \hat{k}$ and $\overrightarrow{\mathrm{AD}}=\hat{i}+2 \hat{j}+\lambda \hat{k}, \lambda \in \mathbb{R}$. Let the projection of the vector $\vec{v}=\hat{i}+\hat{j}+\hat{k}$ on the diagonal $\overrightarrow{\mathrm{AC}}$ of the parallelogram ABCD be of length one unit. If $\alpha, \beta$, where $\alpha>\beta$, be the roots of the equation $\lambda^{2} x^{2}-6 \lambda x+5=0$, then $2 \alpha-\beta$ is equal to
Let $\vec{a}=2 \hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}}, \vec{b}=\hat{\mathrm{i}}+3 \hat{\mathrm{j}}-\hat{\mathrm{k}}$ and $\vec{c}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+3 \hat{\mathrm{k}}$. Let $\vec{v}$ be the vector in the plane of the vectors $\vec{a}$ and $\vec{b}$, such that the length of its projection on the vector $\vec{c}$ is $\frac{1}{\sqrt{14}}$. Then $|\vec{v}|$ is equal to
1966 questions
1411 questions
615 questions
389 questions
224 questions
Work through every JEE Main Vectors & 3D Geometry PYQ, year by year.