TS EAMCET · Maths · Vector Algebra
\(\mathbf{p}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{q}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}\). If the vectors \(\mathbf{a}\) and \(\mathbf{b}\) are the orthogonal projections of \(\mathbf{p}\) on \(\mathbf{q}\) and \(\mathbf{q}\) on \(\mathbf{p}\) respectively, then \(\frac{\mathbf{a} \times \mathbf{b}}{\mathbf{a} \cdot \mathbf{b}}=\)
- A \(\frac{2 \hat{i}+3 \hat{j}+5 \hat{k}}{19 \sqrt{2}}\)
- B \(\frac{2 \hat{i}+3 \hat{j}+5 \hat{k}}{\sqrt{38}}\)
- C \(\frac{2 \hat{i}+3 \hat{j}+5 \hat{k}}{2}\)
- D \(\frac{3 \hat{i}-2 \hat{j}}{13}\)
Answer & Solution
Correct Answer
(C) \(\frac{2 \hat{i}+3 \hat{j}+5 \hat{k}}{2}\)
Step-by-step Solution
Detailed explanation
We have, \(\mathbf{p}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}}, \mathbf{q}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}\)…
See the Complete Solution
Get step-by-step explanations for this and 2.5 Lakh+ more JEE, NEET & CET questions.
- Unlock all solutions
- Practice the full chapter
- Track accuracy across PYQs
4.8 rated on Google Play · 14,000+ reviews
More questions from Maths
- Maximum value of isTS EAMCET 2019 Medium
- Let \(\alpha \neq \beta \quad\) satisfy \(\alpha^2+1=6 \alpha, \beta^2+1=6 \beta\). Then, the quadratic equation whose roots are \(\frac{\alpha}{\alpha+1}, \frac{\beta}{\beta+1}\), isTS EAMCET 2015 Easy
- \(\begin{aligned} & \text { If } 5(f(x))^2=x f(x)+30 \text { and } \ & \int \frac{\left(3 x^3+\left(1-30 x^2\right) f(x)\right)}{(10 f(x)-x)\left(x^3-f(x)\right)^2} d x \ & =\frac{A}{B x^3+D f(x)}+C \text { then } A+B+D=\end{aligned}\)TS EAMCET 2020 Hard
- If the origin is shifted to a point \((h, k)\) by translation of axes in order to make the equation \(x^2+5 x y+2 y^2+5 x+6 y+7=0\) free from first order terms, thenTS EAMCET 2019 Medium
- If \(\alpha\) represents the number of arrangements of \(p\) men and \(q\) women in a row such that all men are together and \(\beta\) represents the number of circular arrangements of the same people with the same condition, then \(\alpha: \beta\) isTS EAMCET 2020 Easy
- If \(A+B+C=2 S\), then \(\sin (S-A) \cos (S-B)-\sin (S-C) \cos S=\)TS EAMCET 2024 Hard
More PYQs from TS EAMCET
- If \(A\) is a square matrix such that \(A(\operatorname{adj} A)=\left[\begin{array}{lll}4 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 4\end{array}\right]\), then \(\operatorname{det}(\operatorname{adj} A)\) is equal toTS EAMCET 2007 Medium
- For anyTS EAMCET 2021 Easy
- Let \(P(1,-2,5)\) be the foot of the perpendicular drawn from the origin to the plane \(\pi_1\) and the same \(P\) be the foot of the perpendicular from \((1,2,-1)\) to the plane \(\pi_2\). Then the acute angle between the planes \(\pi_1\) and \(\pi_2\) isTS EAMCET 2019 Medium
- If \(f(x)=\sin ^6 x+\cos ^6 x+2 \sin ^3 x \cos ^3 x\), then \(\int_0^{\pi / 4} \frac{\sin ^2 2 x}{f(x)} d x=\)TS EAMCET 2020 Medium
- Suppose \(\theta_1\) and \(\theta_2\) are such that \(\left(\theta_1-\theta_2\right)\) lies in \(3^{\text {rd }}\) or \(4^{\text {th }}\) quadrant. If \(\sin \theta_1+\sin \theta_2=-\frac{21}{65}\) and \(\cos \theta_1+\cos \theta_2=-\frac{27}{65}\) then \(\cos \left(\frac{\theta_1-\theta_2}{2}\right)=\)TS EAMCET 2024 Medium
- The slopes of the focal chords of the parabola \(y^2=32 x\), which are tangents to the circle \(x^2+y^2=4\), areTS EAMCET 2014 Hard