Skip to main content
Back to CROSS PRODUCT

JEE MATHEMATICS MCQ: CROSS PRODUCT

1 Medium
1 of 1
Analyzing +4 | -1
A=[l1m1n1l2m2n2l3m3n3]A=\left[\begin{array}{lll}l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2 \\ l_3 & m_3 & n_3\end{array}\right] and B=[p1q1r1p2q2r2p3q3r3]B=\left[\begin{array}{lll}p_1 & q_1 & r_1 \\ p_2 & q_2 & r_2 \\ p_3 & q_3 & r_3\end{array}\right], where
pi,qi,rip_i, q_i, r_i are the cofactors of the elements li,mi,nil_i, m_i, n_i for i=1,2,3i=1,2,3. If (l1,m1,n1),(l2,m2,n2)\left(l_1, m_1, n_1\right),\left(l_2, m_2, n_2\right) and (l3,m3,n3)\left(l_3, m_3, n_3\right) are the direction cosines of three mutually perpendicular lines, then (p1,q1,r1),(p2,q2,r2)\left(p_1, q_1, r_1\right),\left(p_2, q_2, r_2\right) and (p3,q3,r3)\left(p_3, q_3, r_3\right) are
View answer and explanation
Correct option: A
the direction cosines of three mutually perpendicular lines
The correct answer is A (\left{\theta \in(0,2 \pi): \frac{{1+2 i \sin \theta}}{{1-i \sin \theta}} \text { is purely imaginary}\right}.\left{\theta \in(0,2 \pi): \frac{{1+2 i \sin \theta}}{{1-i \sin \theta}} \text { is purely imaginary}\right}.) because the sum of the elements in A is 4π\pi after finding the values of θ\theta that make the expression purely imaginary.

More Questions from CROSS PRODUCT

Practice All Chapter MCQs →
A solution of the vector equation r⃗×b⃗=a⃗×b⃗\vec{r} \times \vec{b}=\vec{a} \times \vec{b}, where a⃗,b⃗\vec{a}, \vec{b} are two given vectors, is where λ\lambda is a parameter. Let a=3i^+2j^+xk^   and  b=i^−j^+k^a = 3\hat i + 2\hat j + x\hat k\mathrm{ }\,\mathrm{\,\,and}\,\,\mathrm{ }b = \hat i - \hat j + \hat k for some real x.x. Then ∣a×b∣=r\left| {a \times b} \right| = r is possible if A tetrahedron has vertices P(1, 2, 1), Q(2, 1, 3), R(-1, 1, 2) and O(0, 0, 0). The angle between the faces OPQ and PQR is Let a=2i^+j^−2k^   and   b=i^+j^.a = 2\widehat i + \widehat j - 2\widehat k{\text{ }}\,\,{\text{and }}\,\,b = \widehat i + \widehat j. If c is a vector such that a⋅c=∣c∣,  ∣c−a∣=22a \cdot c = |c|,\,\,|c - a\mid = 2\sqrt 2 and the angle between a×ba \times b and cc is 30∘,30^\circ , then ∣(a×b)×c∣|(a \times b) \times c| equal to If A⃗=i−2j−3k,B⃗=2i+j−k,C⃗=i+3j−2k\vec{A}=i-2 j-3 k, \vec{B}=2 i+j-k, \vec{C}=i+3 j-2 k, then (A⃗×B⃗)×C⃗(\vec{A} \times \vec{B}) \times \vec{C} is Let OA→=a⃗,OB→=12a⃗+4b⃗\overrightarrow{O A}=\vec{a}, \overrightarrow{O B}=12 \vec{a}+4 \vec{b}, where OO is the origin, If SS is the parallelogram with adjacent sides OAO A and OCO C, then  area of the quadrilateral OABC area of S\frac{\text { area of the quadrilateral } O A B C}{\text { area of } S} is equal to
Type at least 2 characters or jump directly:
🩺 NEET MCQs 🔬 JEE MCQs 🎯 JEE College Predictor Study guides 📊 Dashboard
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.

Continue with Free Account →