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JEE MATHEMATICS MCQ: DOT PRODUCT OF VECTORS

1 Easy
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Application +4 | -1
The plane through the point (βˆ’1,βˆ’1,βˆ’1)\left( { - 1, - 1, - 1} \right) and containing the line of intersection of the planes r⃗  ⋅  (iΛ™^+3 jΛ™^βˆ’k^)=0\mathbf{\vec r}\,\, \cdot \,\,\left( {\widehat {\mathbf{\dot i}} + 3\,\widehat {\mathbf{\dot j}} - \mathbf{\hat k}} \right) = 0 and r⃗  ⋅  (jΛ™^+2 k^)=0\mathbf{\vec r}\,\, \cdot \,\,\left( {\widehat {\mathbf{\dot j}} + 2\,\mathbf{\hat k}} \right) = 0 is
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Correct option: A
r⃗  ⋅  (iΛ™^+2jΛ™^βˆ’3k^)=0\mathbf{\vec r}\,\, \cdot \,\,\left( {\widehat {\mathbf{\dot i}} + 2\widehat {\mathbf{\dot j}} - 3\mathbf{\hat k}} \right) = 0
Option A is correct because the equation of the plane is found by considering the linear combination of the two given planes and using the given point to solve for the constant.

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The distance between the planes given by r.(i+2jβˆ’2k)+5=0\mathbf{r} .(\mathbf{i}+2 \mathbf{j}-2 \mathbf{k})+5=0 and rβˆ’(i+2jβˆ’2k)βˆ’8=0\mathbf{r}-(\mathbf{i}+2 \mathbf{j}-2 \mathbf{k})-8=0 is A vector which makes equal angles with the vectors 13(i^βˆ’2j^+2k^),15(βˆ’4i^βˆ’3k^)\frac{1}{3}\left( {\hat i - 2\hat j + 2\hat k} \right),\frac{1}{5}\left( { - 4\hat i - 3\hat k} \right), and j^{\hat j}, is If vector a=2i^βˆ’3j^+6k^\mathbf{a}=2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}} and vector b=βˆ’2i^+2j^βˆ’k^\mathbf{b}=-2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}, then Β ProjectionΒ ofΒ vectorΒ aΒ onΒ vectorΒ bΒ ProjectionΒ ofΒ vectorΒ bΒ onΒ vectorΒ a\frac{\text { Projection of vector } \mathbf{a} \text { on vector } \mathbf{b}}{\text { Projection of vector } \mathbf{b} \text { on vector } \mathbf{a}} is equal to The angle between aβƒ—{\mathbf{\vec a}} and bβƒ—{\mathbf{\vec b}} is 5Ο€6\frac{{5\pi }}{6} and the projection of aβƒ—{\mathbf{\vec a}} in the direction of bβƒ—{\mathbf{\vec b}} is βˆ’63\frac{{ - 6}}{{\sqrt 3 }} then ∣aβƒ—βˆ£\left| {\mathbf{\vec a}} \right| is equal to A velocity 14Β m/s\frac{1}{4} \mathrm{~m} / \mathrm{s} is resolved into two components along OAO A and OBO B making angles 30∘30^{\circ} and 45∘45^{\circ} respectively with the given velocity. Then the component along OBO B is Find the equation of the perpendicular drown from the origin to the plane 2x+4yβˆ’5z=102x + 4y - 5z = 10
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