Question:
Let \(\bar{a}=\hat{i}+\hat{j}+\hat{k}\) and \(\bar{b}=\hat{i}-\hat{j}+2\hat{k}\). If a vector \(\bar{c}\) is coplanar with \(\bar{a}\) and \(\bar{b}\) such that \(\bar{c}\cdot \bar{a}=1\) and \(\bar{c}\) is perpendicular to \(\bar{b}\), then \(\bar{c}\) is equal to ...
Let \(\bar{a}=\hat{i}+\hat{j}+\hat{k}\) and \(\bar{b}=\hat{i}-\hat{j}+2\hat{k}\). If a vector \(\bar{c}\) is coplanar with \(\bar{a}\) and \(\bar{b}\) such that \(\bar{c}\cdot \bar{a}=1\) and \(\bar{c}\) is perpendicular to \(\bar{b}\), then \(\bar{c}\) is equal to ...
Show Hint
Write c as a combination of a and b, then use c.a = 1 and c.b = 0.
Updated On: Nov 12, 2026
- \(\frac{1}{3}(2\hat{i}+\hat{j}+\hat{k})\)
- \(\frac{1}{5}(2\hat{i}-\hat{j}+4\hat{k})\)
- \(\frac{1}{7}(2\hat{i}+4\hat{j}+\hat{k})\)
- \(\frac{1}{7}(2\hat{i}-4\hat{j}-\hat{k})\)
Show Solution
Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept
Since \(\bar{c}\) is coplanar with \(\bar{a}\) and \(\bar{b}\), write \(\bar{c}=\lambda \bar{a}+\mu \bar{b}\).
Step 2: Dot products
\(\bar{a}\cdot \bar{a}=3\), \(\bar{b}\cdot \bar{b}=6\), \(\bar{a}\cdot \bar{b}=1-1+2=2\).
Step 3: Form equations
\(\bar{c}\cdot \bar{a}=3\lambda +2\mu =1\) and \(\bar{c}\cdot \bar{b}=2\lambda +6\mu =0\).
So \(\lambda =-3\mu\), giving \(-7\mu =1\), \(\mu =-\dfrac{1}{7}\), \(\lambda =\dfrac{3}{7}\).
Step 4: Find c
\(\bar{c}=\dfrac{3}{7}(\hat{i}+\hat{j}+\hat{k})-\dfrac{1}{7}(\hat{i}-\hat{j}+2\hat{k})=\dfrac{1}{7}(2\hat{i}+4\hat{j}+\hat{k})\).
Final Answer:
The vector is \(\dfrac{1}{7}(2\hat{i}+4\hat{j}+\hat{k})\). \[ \boxed{\text{(C) }\dfrac{1}{7}(2\hat{i}+4\hat{j}+\hat{k})} \]
Since \(\bar{c}\) is coplanar with \(\bar{a}\) and \(\bar{b}\), write \(\bar{c}=\lambda \bar{a}+\mu \bar{b}\).
Step 2: Dot products
\(\bar{a}\cdot \bar{a}=3\), \(\bar{b}\cdot \bar{b}=6\), \(\bar{a}\cdot \bar{b}=1-1+2=2\).
Step 3: Form equations
\(\bar{c}\cdot \bar{a}=3\lambda +2\mu =1\) and \(\bar{c}\cdot \bar{b}=2\lambda +6\mu =0\).
So \(\lambda =-3\mu\), giving \(-7\mu =1\), \(\mu =-\dfrac{1}{7}\), \(\lambda =\dfrac{3}{7}\).
Step 4: Find c
\(\bar{c}=\dfrac{3}{7}(\hat{i}+\hat{j}+\hat{k})-\dfrac{1}{7}(\hat{i}-\hat{j}+2\hat{k})=\dfrac{1}{7}(2\hat{i}+4\hat{j}+\hat{k})\).
Final Answer:
The vector is \(\dfrac{1}{7}(2\hat{i}+4\hat{j}+\hat{k})\). \[ \boxed{\text{(C) }\dfrac{1}{7}(2\hat{i}+4\hat{j}+\hat{k})} \]
Was this answer helpful?
0
0
Top MHT CET Mathematics Questions
- The line $5x + y - 1 = 0$ coincides with one of the lines given by $5x^2 + xy - kx - 2y + 2 = 0 $ then the value of k is
- If $\int\frac{f\left(x\right)}{log \left(sin\,x\right)}dx = log\left[log\,sin\,x\right]+c$ then $f\left(x\right)=$
- If $\int\limits^{K}_0 \frac{dx}{2 + 18 x^2} = \frac{\pi}{24}$, then the value of K is
- If $\int^{\pi/2}_{0} \log\cos x dx =\frac{\pi}{2} \log\left(\frac{1}{2}\right)$ then $ \int^{\pi/2}_{0} \log\sec x dx = $
- The point on the curve $y = \sqrt{x - 1}$ where the tangent is perpendicular to the line $2x + y - 5 = 0 $ is
View More Questions
Top MHT CET Product of Two Vectors Questions
- If \( \vec{A} = 2\hat{i} + 3\hat{j} \) and \( \vec{B} = 4\hat{i} - \hat{j} \), then the dot product \( \vec{A} \cdot \vec{B} \) is:
- If \( \mathbf{a} \) and \( \mathbf{b} \) are two non-zero vectors such that the angle between them is \( 60^\circ \), what is the probability that the dot product \( \mathbf{a} \cdot \mathbf{b} \) is positive?
- If \( \vec{a} = 2\hat{i} - \hat{j} + \hat{k} \) and \( \vec{b} = \hat{i} + 2\hat{j} - 3\hat{k} \), find the unit vector perpendicular to both \( \vec{a} \) and \( \vec{b} \).
- Find the angle between non-zero vectors \( \mathbf{a} \) and \( \mathbf{b} \) if their dot product \( \mathbf{a}\cdot\mathbf{b} = 0 \).
- Find the unit vector perpendicular to both \( \vec{a} = 2\hat{i} - \hat{j} + \hat{k} \) and \( \vec{b} = \hat{i} + 2\hat{j} - 3\hat{k} \).
View More Questions
Top MHT CET Questions
- During r- DNA technology, which one of the following enzymes is used for cleaving DNA molecule ?
- In non uniform circular motion, the ratio of tangential to radial acceleration is (r = radius of circle, $v =$ speed of the particle, $\alpha =$ angular acceleration)
- The temperature of $32^{\circ}C$ is equivalent to
- The line $5x + y - 1 = 0$ coincides with one of the lines given by $5x^2 + xy - kx - 2y + 2 = 0 $ then the value of k is
- The heat of formation of water is $ 260\, kJ $ . How much $ H_2O $ is decomposed by $ 130\, kJ $ of heat ?
View More Questions